\documentclass[../libro.tex]{subfiles}

\begin{document}

\ifSubfilesClassLoaded{\appendix\chapter{Afterstories}\clearpage}{}

\section{杂例}

\begin{example}
    设 \(A\) 是 \(n\)~级阵.
    设 \(n\)~级阵 \(D\) 适合
    \begin{align*}
        [D]_{i,j}
        = \begin{cases}
              d_i, & i = j;       \\
              0,   & \text{别的情形}.
          \end{cases}
    \end{align*}
    则
    \begin{align*}
             &
        \det {(A + D)}
        \\
        = {} &
        \hphantom{{} + {}}
        \det {(D)}
        \\
             &
        + \sum_{k = 1}^{n-1}
        {
        \sum_{1 \leq j_1 < j_2 < \dots < j_k \leq n}
        \det {\left(
            A\binom{j_1,\dots,j_k}{j_1,\dots,j_k}
            \right)}
        \det {(D({j_1,\dots,j_k}|{j_1,\dots,j_k}))}
        }
        \\
             &
        + \det {(A)}.
    \end{align*}

    设 \(A\) 的列~\(1\), \(2\), \(\dots\), \(n\)
    分别是 \(b_1^{(1)}\), \(b_2^{(1)}\), \(\dots\), \(b_n^{(1)}\).
    设 \(D\) 的列~\(1\), \(2\), \(\dots\), \(n\)
    分别是 \(b_1^{(0)}\), \(b_2^{(0)}\), \(\dots\), \(b_n^{(0)}\).
    则 \(A + D\) 的列~\(j\) 是
    \begin{align*}
        b_j^{(1)} + b_j^{(0)}
        = b_j^{(0)} + b_j^{(1)}
        = \sum_{c_j = 0}^{1} {b_j^{(c_j)}}.
    \end{align*}
    则, 由多线性,
    \begin{align*}
             &
        \det {(A + D)}
        \\
        = {} &
        \det {\left[
                \sum_{c_1 = 0}^{1} b_1^{(c_1)},
                \sum_{c_2 = 0}^{1} b_2^{(c_2)},
                \dots,
                \sum_{c_n = 0}^{1} b_n^{(c_n)}
                \right]}
        \\
        = {} &
        \sum_{c_1 = 0}^{1}
        \det {\left[
                b_1^{(c_1)},
                \sum_{c_2 = 0}^{1} b_2^{(c_2)},
                \dots,
                \sum_{c_n = 0}^{1} b_n^{(c_n)}
                \right]}
        \\
        = {} &
        \sum_{c_1 = 0}^{1}
        \sum_{c_2 = 0}^{1}
        \det {\left[
                b_1^{(c_1)},
                b_2^{(c_2)},
                \dots,
                \sum_{c_n = 0}^{1} b_n^{(c_n)}
                \right]}
        \\
        = {} &
        \dots
        \\
        = {} &
        \sum_{c_1 = 0}^{1}
        \sum_{c_2 = 0}^{1}
        \dots
        \sum_{c_n = 0}^{1}
        \det {[b_1^{(c_1)}, b_2^{(c_2)}, \dots, b_n^{(c_n)}]}
        \\
        = {} &
        \sum_{0 \leq c_1, c_2, \dots, c_n \leq 1}
        \det {[b_1^{(c_1)}, b_2^{(c_2)}, \dots, b_n^{(c_n)}]}.
    \end{align*}
    由加法的结合律与交换律,
    我们可按任何方式, 任何次序求这些
    \begin{align*}
        \det {[b_1^{(c_1)}, b_2^{(c_2)}, \dots, b_n^{(c_n)}]}
    \end{align*}
    的和.
    特别地, 我们可按 \(c_1 + c_2 + \dots + c_n\) 分%
    这些数为若干组,
    求出一组的组和 (即一组的元的和),
    再求这些组的组和的和.
    因为 \(0 \leq c_j \leq 1\),
    故 \(0 \leq c_1 + c_2 + \dots + c_n \leq n\).
    于是, 我们可分这些数为 \(n+1\)~组:
    适合 \(c_1 + c_2 + \dots + c_n = 0\) 的项在一组;
    适合 \(c_1 + c_2 + \dots + c_n = 1\) 的项在一组;
    ……
    适合 \(c_1 + c_2 + \dots + c_n = n\) 的项在一组.
    不难看出,
    每项在某一组里,
    且每项不能在二个不同的组里.
    则
    \begin{align*}
             &
        \det {(A + D)}
        \\
        = {} &
        \sum_{0 \leq c_1, c_2, \dots, c_n \leq 1}
        \det {[b_1^{(c_1)}, b_2^{(c_2)}, \dots, b_n^{(c_n)}]}
        \\
        = {} &
        \sum_{k = 0}^{n}\,
        \sum_{\substack{0 \leq c_1, c_2, \dots, c_n \leq 1 \\
                c_1 + c_2 + \dots + c_n = k}}
        \det {[b_1^{(c_1)}, b_2^{(c_2)}, \dots, b_n^{(c_n)}]}
        \\
        = {} &
        \hphantom{{} + {}}
        \sum_{\substack{0 \leq c_1, c_2, \dots, c_n \leq 1 \\
                c_1 + c_2 + \dots + c_n = 0}}
        \det {[b_1^{(c_1)}, b_2^{(c_2)}, \dots, b_n^{(c_n)}]}
        \\
             &
        +
        \sum_{k = 1}^{n-1}\,
        \sum_{\substack{0 \leq c_1, c_2, \dots, c_n \leq 1 \\
                c_1 + c_2 + \dots + c_n = k}}
        \det {[b_1^{(c_1)}, b_2^{(c_2)}, \dots, b_n^{(c_n)}]}
        \\
             &
        +
        \sum_{\substack{0 \leq c_1, c_2, \dots, c_n \leq 1 \\
                c_1 + c_2 + \dots + c_n = n}}
        \det {[b_1^{(c_1)}, b_2^{(c_2)}, \dots, b_n^{(c_n)}]}.
    \end{align*}

    不难看出, \(c_1 + c_2 + \dots + c_n = 0\) 时,
    \(c_1 = c_2 = \dots = c_n = 0\).
    则
    \begin{align*}
        \sum_{0 \leq c_1, c_2, \dots, c_n \leq 1}
        \det {[b_1^{(c_1)}, b_2^{(c_2)}, \dots, b_n^{(c_n)}]}
        =
        \det {[b_1^{(0)}, b_2^{(0)}, \dots, b_n^{(0)}]}
        =
        \det {(D)}.
    \end{align*}

    不难看出, \(c_1 + c_2 + \dots + c_n = n\) 时,
    \(c_1 = c_2 = \dots = c_n = 1\).
    则
    \begin{align*}
        \sum_{0 \leq c_1, c_2, \dots, c_n \leq 1}
        \det {[b_1^{(c_1)}, b_2^{(c_2)}, \dots, b_n^{(c_n)}]}
        =
        \det {[b_1^{(1)}, b_2^{(1)}, \dots, b_n^{(1)}]}
        =
        \det {(A)}.
    \end{align*}

    设正整数 \(k < n\).
    由 \(c_1 + c_2 + \dots + c_n = k\),
    知在 \(c_1\), \(c_2\), \(\dots\), \(c_n\)
    中, 有 \(k\)~个 \(1\) 与 \(n-k\)~个 \(0\).
    设 \(c_{j_1} = c_{j_2} = \dots = c_{j_k} = 1\)
    (其中, \(1 \leq j_1 < j_2 < \dots < j_k \leq n\)),
    且 \(c_{j_{k+1}} = \dots = c_{j_n} = 0\)
    (其中, \(1 \leq j_{k+1} < \dots < j_n \leq n\)).
    记 \(n\)~级阵
    \begin{align*}
        B_{c_1,\dots,c_n}
        = [b_1^{(c_1)}, b_2^{(c_2)}, \dots, b_n^{(c_n)}].
    \end{align*}
    注意, 当 \(\ell > k\),
    且 \(i \neq j_\ell\) 时,
    \([B_{c_1,\dots,c_n}]_{i,j_\ell} = 0\),
    且 \([B_{c_1,\dots,c_n}]_{j_\ell,j_\ell} = d_{j_\ell}\),
    按列~\(j_{k+1}\) 展开 \(\det {(B_{c_1,\dots,c_n})}\),
    有
    \begin{align*}
        \det {(B_{c_1,\dots,c_n})}
        = {} & (-1)^{j_{k+1}+j_{k+1}}
            [B_{c_1,\dots,c_n}]_{j_{k+1},j_{k+1}}
        \det {(B_{c_1,\dots,c_n}(j_{k+1}|j_{k+1}))}
        \\
        = {} &
        d_{j_{k+1}} \det {(B_{c_1,\dots,c_n}(j_{k+1}|j_{k+1}))}.
    \end{align*}
    按列~\(j_{k+2}-1\) 展开
    \(\det {(B_{c_1,\dots,c_n}(j_{k+1}|j_{k+1}))}\),
    有
    \begin{align*}
             &
        \det {(B_{c_1,\dots,c_n}(j_{k+1}|j_{k+1}))}
        \\
        = {} & (-1)^{j_{k+2}-1+j_{k+2}-1}
            [B_{c_1,\dots,c_n}]_{j_{k+2},j_{k+2}}
        \det {(B_{c_1,\dots,c_n}
        ({j_{k+1},j_{k+2}}|{j_{k+1},j_{k+2}}))}
        \\
        = {} &
        d_{j_{k+2}}
        \det {(B_{c_1,\dots,c_n}
        ({j_{k+1},j_{k+2}}|{j_{k+1},j_{k+2}}))}.
    \end{align*}
    故
    \begin{align*}
        \det {(B_{c_1,\dots,c_n})}
        = d_{j_{k+1}} d_{j_{k+2}}
        \det {(B_{c_1,\dots,c_n}
        ({j_{k+1},j_{k+2}}|{j_{k+1},j_{k+2}}))}.
    \end{align*}
    ……
    最后, 我们算出
    \begin{align*}
             &
        \det {(B_{c_1,\dots,c_n})}
        \\
        = {} &
        d_{j_{k+1}} d_{j_{k+2}} \dots d_{j_n}
        \det {(B_{c_1,\dots,c_n}
        ({j_{k+1},\dots,j_n}|{j_{k+1},\dots,j_n}))}
        \\
        = {} &
        \det {(B_{c_1,\dots,c_n}
        ({j_{k+1},\dots,j_n}|{j_{k+1},\dots,j_n}))}
        \,
        d_{j_{k+1}} d_{j_{k+2}} \dots d_{j_n}
        \\
        = {} &
        \det {\left(
            A\binom{j_1,\dots,j_k}{j_1,\dots,j_k}
            \right)}
        \det {(D({j_1,\dots,j_k}|{j_1,\dots,j_k}))}.
    \end{align*}
    于是
    \begin{align*}
             &
        \sum_{\substack{0 \leq c_1, c_2, \dots, c_n \leq 1 \\
                c_1 + c_2 + \dots + c_n = k}}
        \det {[b_1^{(c_1)}, b_2^{(c_2)}, \dots, b_n^{(c_n)}]}
        \\
        = {} &
        \sum_{1 \leq j_1 < j_2 < \dots < j_k \leq n}
        \det {\left(
            A\binom{j_1,\dots,j_k}{j_1,\dots,j_k}
            \right)}
        \det {(D({j_1,\dots,j_k}|{j_1,\dots,j_k}))}.
    \end{align*}

    综上,
    \begin{align*}
             &
        \det {(A + D)}
        \\
        = {} &
        \hphantom{{} + {}}
        \sum_{\substack{0 \leq c_1, c_2, \dots, c_n \leq 1 \\
                c_1 + c_2 + \dots + c_n = 0}}
        \det {[b_1^{(c_1)}, b_2^{(c_2)}, \dots, b_n^{(c_n)}]}
        \\
             &
        +
        \sum_{k = 1}^{n-1}\,
        \sum_{\substack{0 \leq c_1, c_2, \dots, c_n \leq 1 \\
                c_1 + c_2 + \dots + c_n = k}}
        \det {[b_1^{(c_1)}, b_2^{(c_2)}, \dots, b_n^{(c_n)}]}
        \\
             &
        +
        \sum_{\substack{0 \leq c_1, c_2, \dots, c_n \leq 1 \\
                c_1 + c_2 + \dots + c_n = n}}
        \det {[b_1^{(c_1)}, b_2^{(c_2)}, \dots, b_n^{(c_n)}]}
        \\
        = {} &
        \hphantom{{} + {}}
        \det {(D)}
        \\
             &
        + \sum_{k = 1}^{n-1}
        {
        \sum_{1 \leq j_1 < j_2 < \dots < j_k \leq n}
        \det {\left(
            A\binom{j_1,\dots,j_k}{j_1,\dots,j_k}
            \right)}
        \det {(D({j_1,\dots,j_k}|{j_1,\dots,j_k}))}
        }
        \\
             &
        + \det {(A)}.
    \end{align*}
\end{example}

\begin{example}
    设 \(A\) 是 \(n\)~级阵.
    设 \(x\) 是数.
    则
    \begin{align*}
        %  &
        \det {(xI_n + A)}
        % \\
        =
        % {} &
        x^n
        + \sum_{k = 1}^{n}
            {x^{n-k}
                \sum_{1 \leq j_1 < \dots < j_k \leq n}
                \det {\left(
                    A\binom{j_1,\dots,j_k}{j_1,\dots,j_k}
                    \right)}
            }.
    \end{align*}

    注意,
    \begin{align*}
        [xI_n]_{i,j}
        = \begin{cases}
              x, & i = j;       \\
              0, & \text{别的情形}.
          \end{cases}
    \end{align*}
    故, 由上个例,
    \begin{align*}
             &
        \det {(xI_n + A)}
        \\
        = {} &
        \hphantom{{} + {}}
        \det {(xI_n)}
        \\
             &
        + \sum_{k = 1}^{n-1}
        {
        \sum_{1 \leq j_1 < \dots < j_k \leq n}
        \det {\left(
            A\binom{j_1,\dots,j_k}{j_1,\dots,j_k}
            \right)}
        \det {((xI_n)({j_1,\dots,j_k}|{j_1,\dots,j_k}))}
        }
        \\
             &
        + \det {(A)}
        \\
        = {} &
        \hphantom{{} + {}}
        x^n
        % \\
        %      &
        + \sum_{k = 1}^{n-1}
            {
                \sum_{1 \leq j_1 < \dots < j_k \leq n}
                \det {\left(
                    A\binom{j_1,\dots,j_k}{j_1,\dots,j_k}
                    \right)}
                \,
                x^{n-k}
            }
        \\
             &
        + \det {(A)}
        \\
        = {} &
        \hphantom{{} + {}}
        x^n
        % \\
        %      &
        + \sum_{k = 1}^{n-1}
            {
                x^{n-k}
                \sum_{1 \leq j_1 < \dots < j_k \leq n}
                \det {\left(
                    A\binom{j_1,\dots,j_k}{j_1,\dots,j_k}
                    \right)}
            }
        \\
             &
        + x^{n-n}
        \sum_{1 \leq j_1 < \dots < j_n \leq n}
        \det {\left(
            A\binom{j_1,\dots,j_n}{j_1,\dots,j_n}
            \right)}
        \\
        = {} &
        x^n
        + \sum_{k = 1}^{n}
            {x^{n-k}
                \sum_{1 \leq j_1 < \dots < j_k \leq n}
                \det {\left(
                    A\binom{j_1,\dots,j_k}{j_1,\dots,j_k}
                    \right)}
            }.
    \end{align*}
\end{example}

\begin{example}\label{emp:sum-of-principal-minors-of-AB-and-that-of-BA}
    设正整数 \(m\), \(n\) 适合 \(m \geq n\).
    设 \(A\), \(B\) 分别是 \(m \times n\) 与 \(n \times m\)~阵.
    设正整数 \(k \leq n\).
    由 Binet--Cauchy 公式的推广,
    \begin{align*}
             &
        \sum_{1 \leq j_1 < \dots < j_k \leq m}
        \det {\left(
            (AB)\binom{j_1,\dots,j_k}{j_1,\dots,j_k}
            \right)}
        \\
        = {} &
        \sum_{1 \leq j_1 < \dots < j_k \leq m}
        \sum_{1 \leq i_1 < \dots < i_k \leq n}
        \det {\left(
            A\binom{j_1,\dots,j_k}{i_1,\dots,i_k}
            \right)}
        \det {\left(
            B\binom{i_1,\dots,i_k}{j_1,\dots,j_k}
            \right)}
        \\
        = {} &
        \sum_{1 \leq j_1 < \dots < j_k \leq m}
        \sum_{1 \leq i_1 < \dots < i_k \leq n}
        \det {\left(
            B\binom{i_1,\dots,i_k}{j_1,\dots,j_k}
            \right)}
        \det {\left(
            A\binom{j_1,\dots,j_k}{i_1,\dots,i_k}
            \right)}
        \\
        = {} &
        \sum_{1 \leq i_1 < \dots < i_k \leq n}
        \sum_{1 \leq j_1 < \dots < j_k \leq m}
        \det {\left(
            B\binom{i_1,\dots,i_k}{j_1,\dots,j_k}
            \right)}
        \det {\left(
            A\binom{j_1,\dots,j_k}{i_1,\dots,i_k}
            \right)}
        \\
        = {} &
        \sum_{1 \leq i_1 < \dots < i_k \leq n}
        \det {\left(
            (BA)\binom{i_1,\dots,i_k}{i_1,\dots,i_k}
            \right)}.
    \end{align*}
\end{example}

\begin{example}
    设正整数 \(m\), \(n\) 适合 \(m \geq n\).
    设 \(A\), \(B\) 分别是 \(m \times n\) 与 \(n \times m\)~阵.
    设 \(x\) 是数.
    则
    \begin{align*}
             &
        \det {(xI_m + AB)}
        \\
        = {} &
        x^m
        + \sum_{k = 1}^{m}
            {x^{m-k}
                \sum_{1 \leq j_1 < \dots < j_k \leq m}
                \det {\left(
                    (AB)\binom{j_1,\dots,j_k}{j_1,\dots,j_k}
                    \right)}
            }
        \\
        = {} &
        \hphantom{{} + {}}
        x^m
        + \sum_{k = 1}^{n}
            {x^{m-k}
                \sum_{1 \leq j_1 < \dots < j_k \leq m}
                \det {\left(
                    (AB)\binom{j_1,\dots,j_k}{j_1,\dots,j_k}
                    \right)}
            }
        \\
             &
        + \sum_{k = n+1}^{m}
            {x^{m-k}
                \sum_{1 \leq j_1 < \dots < j_k \leq m}
                \det {\left(
                    (AB)\binom{j_1,\dots,j_k}{j_1,\dots,j_k}
                    \right)}
            }
        \\
        = {} &
        \hphantom{{} + {}}
        x^m
        + \sum_{k = 1}^{n}
            {x^{m-k}
                \sum_{1 \leq j_1 < \dots < j_k \leq m}
                \det {\left(
                    (AB)\binom{j_1,\dots,j_k}{j_1,\dots,j_k}
                    \right)}
            }
        \\
             &
        + \sum_{k = n+1}^{m}
            {x^{m-k}
                \sum_{1 \leq j_1 < \dots < j_k \leq m}
                0
            }
        \\
        = {} &
        x^m
        + \sum_{k = 1}^{n}
            {x^{m-k}
                \sum_{1 \leq j_1 < \dots < j_k \leq m}
                \det {\left(
                    (AB)\binom{j_1,\dots,j_k}{j_1,\dots,j_k}
                    \right)}
            }
        \\
        = {} &
        x^{m-n} x^n
        + \sum_{k = 1}^{n}
            {x^{m-n} x^{n-k}
                \sum_{1 \leq i_1 < \dots < i_k \leq n}
                \det {\left(
                    (BA)\binom{i_1,\dots,i_k}{i_1,\dots,i_k}
                    \right)}
            }
        \\
        = {} &
        x^{m-n} x^n
        + x^{m-n} \sum_{k = 1}^{n}
            {x^{n-k}
                \sum_{1 \leq i_1 < \dots < i_k \leq n}
                \det {\left(
                    (BA)\binom{i_1,\dots,i_k}{i_1,\dots,i_k}
                    \right)}
            }
        \\
        = {} &
        x^{m-n}
        \left(
        x^n
        + \sum_{k = 1}^{n}
        {x^{n-k}
            \sum_{1 \leq i_1 < \dots < i_k \leq n}
            \det {\left(
                (BA)\binom{i_1,\dots,i_k}{i_1,\dots,i_k}
                \right)}
        }
        \right)
        \\
        = {} &
        x^{m-n} \det {(xI_n + BA)}.
    \end{align*}
    特别地, 代 \(x\) 以数 \(1\), 有
    \begin{align*}
        \det {(I_m + AB)} = \det {(I_n + BA)}.
    \end{align*}
\end{example}

\begin{example}
    设 \(a_1\), \(b_1\), \(a_2\), \(b_2\),
    \(\dots\), \(a_m\), \(b_m\) 是 \(2m\)~个数.
    作 \(m\)~级阵 \(C\) 如下:
    \begin{align*}
        [C]_{i,j} =
        \begin{cases}
            1 + a_i b_i, & i = j;       \\
            a_i b_j,     & \text{别的情形}.
        \end{cases}
    \end{align*}
    我们计算 \(\det {(C)}\).

    设 \(A = [a_1, a_2, \dots, a_m]^{\mathrm{T}}\),
    且 \(B = [b_1, b_2, \dots, b_m]\).
    则
    \begin{align*}
        [AB]_{i,j} = [A]_{i,1} [B]_{1,j} = a_i b_j.
    \end{align*}
    由此, 不难看出, \(C = I_m + AB\).
    故
    \begin{align*}
        \det {(C)}
        = {} & \det {(I_m + AB)}                        \\
        = {} & \det {(I_1 + BA)}                        \\
        = {} & [I_1 + BA]_{1,1}                         \\
        = {} & [I_1]_{1,1} + [BA]_{1,1}                 \\
        = {} & 1 + b_1 a_1 + b_2 a_2 + \dots + b_m a_m.
    \end{align*}

    我们当然也可用别的方法.
    作 \(m+1\)~级阵 \(G\) 如下:
    \begin{align*}
        [G]_{i,j} =
        \begin{cases}
            1,         & i = j = m+1; \\
            -a_i,      & i < j = m+1; \\
            0,         & m+1 = i > j; \\
            [C]_{i,j}, & \text{别的情形}.
        \end{cases}
    \end{align*}
    通俗地,
    \begin{align*}
        G =
        \begin{bmatrix}
            1 + a_1 b_1 & a_1 b_2     & \cdots & a_1 b_{m-1}         & a_1 b_m     & -a_1     \\
            a_2 b_1     & 1 + a_2 b_2 & \cdots & a_2 b_{m-1}         & a_2 b_m     & -a_2     \\
            \vdots      & \vdots      & {}     & \vdots              & \vdots      & \vdots   \\
            a_{m-1} b_1 & a_{m-1} b_2 & \cdots & 1 + a_{m-1} b_{m-1} & a_{m-1} b_m & -a_{m-1} \\
            a_m b_1     & a_m b_2     & \cdots & a_m b_{m-1}         & 1 + a_m b_m & -a_m     \\
            0           & 0           & \cdots & 0                   & 0           & 1        \\
        \end{bmatrix}.
    \end{align*}
    一方面, 按行~\(m+1\) 展开, 有
    \begin{align*}
        \det {(G)}
        = (-1)^{m+1+m+1}\,1\,\det {(G(1|1))}
        = \det {(C)}.
    \end{align*}
    另一方面, 我们%
    加列~\(m+1\) 的 \(b_1\)~倍于列~\(1\),
    加列~\(m+1\) 的 \(b_2\)~倍于列~\(2\),
    ……
    加列~\(m+1\) 的 \(b_m\)~倍于列~\(m\),
    得 \(m+1\)~级阵
    \begin{align*}
        H =
        \begin{bmatrix}
            1      & 0      & \cdots & 0       & 0      & -a_1     \\
            0      & 1      & \cdots & 0       & 0      & -a_2     \\
            \vdots & \vdots & {}     & \vdots  & \vdots & \vdots   \\
            0      & 0      & \cdots & 1       & 0      & -a_{m-1} \\
            0      & 0      & \cdots & 0       & 1      & -a_m     \\
            b_1    & b_2    & \cdots & b_{m-1} & b_m    & 1        \\
        \end{bmatrix}.
    \end{align*}
    则 \(\det {(H)} = \det {(G)}\).
    故 \(\det {(C)} = \det {(H)}\).

    我们%
    加列~\(1\) 的 \(a_1\)~倍于列~\(m+1\),
    加列~\(1\) 的 \(a_2\)~倍于列~\(m+1\),
    ……
    加列~\(1\) 的 \(a_m\)~倍于列~\(m+1\),
    得 \(m+1\)~级阵
    \begin{align*}
        J =
        \begin{bmatrix}
            1      & 0      & \cdots & 0       & 0      & 0                                       \\
            0      & 1      & \cdots & 0       & 0      & 0                                       \\
            \vdots & \vdots & {}     & \vdots  & \vdots & \vdots                                  \\
            0      & 0      & \cdots & 1       & 0      & 0                                       \\
            0      & 0      & \cdots & 0       & 1      & 0                                       \\
            b_1    & b_2    & \cdots & b_{m-1} & b_m    & 1 + b_1 a_1 + b_2 a_2 + \dots + b_m a_m \\
        \end{bmatrix}.
    \end{align*}
    则 \(\det {(J)} = \det {(H)}\).
    故
    \begin{align*}
        \det {(C)} = \det {(J)}
        = 1 + b_1 a_1 + b_2 a_2 + \dots + b_m a_m.
    \end{align*}
\end{example}

\begin{example}
    设 \(A\) 是 \(n\)~级阵.
    则 \(AA\) 是有意义的, 且也是方阵.
    我们写 \(A^2 = AA\).
    类似地, \(A^2 A\) 也是有意义的, 且也是方阵.
    我们写 \(A^3 = A^2 A\).
    一般地, 对任何非负整数 \(m\),
    我们如下定义方阵 \(A\) 的 \(m\)~次方:
    \begin{align*}
        A^m =
        \begin{cases}
            I,         & m = 0;    \\
            A^{m-1} A, & m \geq 1.
        \end{cases}
    \end{align*}
    不难看出, \(A^1 = A^0 A = I A = A\).

    类似数的非负整数次方, 方阵的非负整数次方适合,
    对任何方阵 \(A\), 任何非负整数 \(\ell\), \(m\),

    (1)
    \(A^{\ell+m} = A^{\ell} A^{m}\);

    (2)
    \((A^{\ell})^{m} = A^{\ell m}\).

    我们用数学归纳法证明 (1) 与 (2).

    (1)
    作命题 \(P(m)\):
    \(A^{\ell+m} = A^{\ell} A^{m}\).
    我们的目标是,
    对任何非负整数 \(m\), \(P(m)\) 是对的.

    \(P(0)\) 是对的, 因为
    \(A^{\ell+0} = A^{\ell} = A^{\ell} I = A^{\ell} A^{0}\).

    假定 \(P(m)\) 是对的.
    我们由此证明, \(P(m+1)\) 也是对的.
    注意,
    \begin{align*}
        A^{\ell+(m+1)}
        = A^{(\ell+m)+1}
        = A^{\ell+m} A
        = (A^{\ell} A^{m}) A
        = A^{\ell} (A^{m} A)
        = A^{\ell} A^{m+1}.
    \end{align*}
    所以, \(P(m+1)\) 是对的.
    由数学归纳法, 待证命题成立.

    (2)
    作命题 \(Q(m)\):
    \((A^{\ell})^{m} = A^{\ell m}\).
    我们的目标是,
    对任何非负整数 \(m\), \(Q(m)\) 是对的.

    \(Q(0)\) 是对的, 因为
    \((A^{\ell})^{0} = I = A^{0} = A^{\ell 0}\).

    假定 \(Q(m)\) 是对的.
    我们由此证明, \(Q(m+1)\) 也是对的.
    注意,
    \begin{align*}
        (A^{\ell})^{m+1}
        = (A^{\ell})^{m} A^{\ell}
        = A^{\ell m} A^{\ell}
        = A^{\ell m + \ell}
        = A^{\ell (m+1)}.
    \end{align*}
    所以, \(Q(m+1)\) 是对的.
    由数学归纳法, 待证命题成立.
\end{example}

\begin{example}
    我们知道, 对任何数 \(x\), \(y\), 任何非负整数 \(m\),
    必 \((xy)^m = x^m y^m\).
    可是, 存在二个同级的方阵 \(A\), \(B\), 存在非负整数 \(m\),
    使 \((AB)^m \neq A^m B^m\):
    取 \(m = 2\),
    \begin{align*}
        A = \begin{bmatrix}
                1 & 1 \\ 0 & 1 \\
            \end{bmatrix},
        \quad
        B = \begin{bmatrix}
                1 & 0 \\ 1 & 1 \\
            \end{bmatrix}.
    \end{align*}
    不难算出
    \begin{align*}
        AB = \begin{bmatrix}
                 2 & 1 \\ 1 & 1 \\
             \end{bmatrix},
        \quad
        A^2 = \begin{bmatrix}
                  1 & 2 \\ 0 & 1 \\
              \end{bmatrix},
        \quad
        B^2 = \begin{bmatrix}
                  1 & 0 \\ 2 & 1 \\
              \end{bmatrix}.
    \end{align*}
    则
    \begin{align*}
        (AB)^2 = \begin{bmatrix}
                     5 & 3 \\ 3 & 2 \\
                 \end{bmatrix},
        \quad
        A^2 B^2 = \begin{bmatrix}
                      5 & 2 \\ 2 & 1 \\
                  \end{bmatrix}.
    \end{align*}

    不过, 若 \(AB = BA\), 则我们仍有,
    对任何非负整数 \(m\), 必 \((AB)^m = A^m B^m\).
    为证明此事, 我们要作准备.

    作命题 \(P(m)\): \(B^{m} A = A B^{m}\).
    (注意, \(A B^{m}\) 是 \(A (B^{m})\),
    而不是 \((AB)^m\);
    这就像 \(x y^{m}\) 是 \(x (y^m)\),
    而不是 \((x y)^m\).)
    我们用数学归纳法证明 \(P(m)\).

    \(P(0)\) 是对的, 因为
    \(B^{0} A = I A = A I = A B^{0}\).

    假定 \(P(m)\) 是对的.
    我们由此证明, \(P(m+1)\) 也是对的.
    注意,
    \begin{align*}
        B^{m+1} A
        = {} &
        (B^{m} B) A
        = B^{m} (B A)
        = B^{m} (A B)
        \\
        = {} &
        (B^{m} A) B
        = (A B^{m}) B
        = A (B^{m} B)
        = A B^{m+1}.
    \end{align*}
    所以, \(P(m+1)\) 是对的.
    由数学归纳法, 待证命题成立.

    好的.
    我们已好地准备.

    作命题 \(Q(m)\):
    \((AB)^{m} = A^{m} B^{m}\).
    我们的目标是,
    对任何非负整数 \(m\), \(Q(m)\) 是对的.

    \(Q(0)\) 是对的, 因为
    \((AB)^0 = I = I I = A^0 B^0\).

    假定 \(Q(m)\) 是对的.
    我们由此证明, \(Q(m+1)\) 也是对的.
    注意,
    \begin{align*}
        (AB)^{m+1}
        = {} &
        (AB)^{m} (AB)
        = (A^{m} B^{m}) (A B)
        = ((A^{m} B^{m}) A) B
        = (A^{m} (B^{m} A)) B
        \\
        = {} &
        (A^{m} (A B^{m})) B
        = ((A^{m} A) B^{m}) B
        = A^{m+1} (B^{m} B)
        = A^{m+1} B^{m+1}.
    \end{align*}
    所以, \(Q(m+1)\) 是对的.
    由数学归纳法, 待证命题成立.
\end{example}

\begin{example}
    设 \(A\) 是方阵.
    设 \(k\) 是数.
    设 \(m\) 是非负整数.
    可以用数学归纳法证明:

    (1)
    \((kA)^m = k^m A^m\).
    (用 \((k A) (\ell B) = (k \ell) (A B)\),
    若 \(A\), \(B\) 都是 \(n\)~级阵,
    且 \(k\), \(\ell\) 是数.)

    (2)
    \(\det {(A^m)} = (\det {(A)})^m\).
    (用 \(\det {(AB)} = \det {(A)} \det {(B)}\),
    若 \(A\), \(B\) 都是 \(n\)~级阵.)

    (3)
    \((A^m)^{\mathrm{T}} = (A^{\mathrm{T}})^m\).
    (用 \((BA)^{\mathrm{T}} = A^{\mathrm{T}} B^{\mathrm{T}}\),
    若 \(A\), \(B\) 都是 \(n\)~级阵;
    见 ``转置的性质''.)

    (4)
    \(\operatorname{adj} {(A^m)}
    = (\operatorname{adj} {(A)})^m\).
    (用 \(\operatorname{adj} {(BA)}
    = \operatorname{adj} {(A)} \operatorname{adj} {(B)}\),
    若 \(A\), \(B\) 都是 \(n\)~级阵;
    见 ``古伴的性质 (3)''.)

    请允许我留它们为您的习题.
\end{example}

\begin{example}
    设 \(A\), \(B\) 是 \(n\)~级阵.
    则存在跟 \(A\), \(B\) 的元有关的数
    \(d_0\), \(d_1\), \(\dots\), \(d_n\),
    使对任何数 \(x\), 有
    \begin{align*}
        \det {(xA + B)}
        = \sum_{k = 0}^{n} {d_k x^{n-k}}
        = d_0 x^n + d_1 x^{n-1} + \dots + d_n,
    \end{align*}
    其中, \(d_0 = \det {(A)}\),
    且 \(d_n = \det {(B)}\).
    % 我们代 \(x\) 以 \(0\), 得
    % \begin{align*}
    %     \det {(0A + B)} = d_0 0^n + d_1 0^{n-1} + \dots + d_n,
    % \end{align*}
    % 即 \(d_n = \det {(B)}\).

    作命题 \(P(m)\):
    对任何 \(m\)~级阵 \(A\), \(B\),
    存在跟 \(A\), \(B\) 的元有关的数
    \(d_0\), \(d_1\), \(\dots\), \(d_m\),
    使对任何数 \(x\), 有
    \begin{align*}
        \det {(xA + B)}
        = \sum_{k = 0}^{m} {d_k x^{m-k}}
        = d_0 x^m + d_1 x^{m-1} + \dots + d_m,
    \end{align*}
    其中, \(d_0 = \det {(A)}\),
    且 \(d_m = \det {(B)}\).
    我们的目标是,
    对任何非负整数 \(m\), \(P(m)\) 是对的.

    \(P(1)\) 是显然的:
    \begin{align*}
        \det {(xA + B)}
        = [xA + B]_{1,1}
        = x [A]_{1,1} + [B]_{1,1}
        = \det {(A)}\, x + \det {(B)}.
    \end{align*}

    假定 \(P(m-1)\) 是对的.
    我们由此证明, \(P(m)\) 也是对的.

    设 \(A\), \(B\) 是 \(m\)~级阵.
    则
    \begin{align*}
        \det {(xA + B)}
        = {} &
        \sum_{i = 1}^{m}
        (-1)^{i+1} [xA + B]_{i,1} \det {((xA + B)(i|1))}
        \\
        = {} &
        \sum_{i = 1}^{m}
        (-1)^{i+1} (x [A]_{i,1} + [B]_{i,1})
        \det {(x A(i|1) + B(i|1))}.
    \end{align*}
    \(A(i|1)\), \(B(i|1)\) 是 \(m-1\)~级阵.
    由假定, 存在跟 \(A(i|1)\), \(B(i|1)\) 的元有关的数
    (从而也是跟 \(A\), \(B\) 的元有关的数)
    \(c_{i,0}\), \(c_{i,1}\), \(\dots\), \(c_{i,m-1}\),
    使对任何数 \(x\), 有
    \begin{align*}
        \det {(x A(i|1) + B(i|1))}
        = \sum_{\ell = 0}^{m-1}
        c_{i,\ell} x^{m-1-\ell},
    \end{align*}
    其中, \(c_{i,0} = \det {(A(i|1))}\),
    且 \(c_{i,m-1} = \det {(B(i|1))}\).
    为方便, 记 \(a_i = (-1)^{i+1} [A]_{i,1}\),
    且 \(b_i = (-1)^{i+1} [B]_{i,1}\).
    则 \(a_i\), \(b_i\) 跟 \(A\), \(B\) 的元有关.
    则
    \begin{align*}
             &
        \det {(xA + B)}
        \\
        = {} &
        \sum_{i = 1}^{m}
        (a_i x + b_i)
        \sum_{\ell = 0}^{m-1}
        c_{i,\ell} x^{m-1-\ell}
        \\
        = {} &
        \sum_{i = 1}^{m}
        \sum_{\ell = 0}^{m-1}
        (a_i x + b_i)
        c_{i,\ell} x^{m-1-\ell}
        \\
        = {} &
        \sum_{i = 1}^{m}
        \sum_{\ell = 0}^{m-1}
        (a_i c_{i,\ell} x^{m-\ell}
        + b_i c_{i,\ell} x^{m-1-\ell})
        \\
        = {} &
        \sum_{\ell = 0}^{m-1}
        \sum_{i = 1}^{m}
        (a_i c_{i,\ell} x^{m-\ell}
        + b_i c_{i,\ell} x^{m-1-\ell})
        \\
        = {} &
        \sum_{\ell = 0}^{m-1}
        \sum_{i = 1}^{m}
        a_i c_{i,\ell} x^{m-\ell}
        +
        \sum_{\ell = 0}^{m-1}
        \sum_{i = 1}^{m}
        b_i c_{i,\ell} x^{m-1-\ell}
        \\
        = {} & \hphantom{{} + {}}
        \sum_{i = 1}^{m}
        a_i c_{i,0} x^{m-0}
        +
        \sum_{\ell = 1}^{m-1}
        \sum_{i = 1}^{m}
        a_i c_{i,\ell} x^{m-\ell}
        \\
             &
        +
        \sum_{\ell = 0}^{m-2}
        \sum_{i = 1}^{m}
        b_i c_{i,\ell} x^{m-1-\ell}
        +
        \sum_{i = 1}^{m}
        b_i c_{i,m-1} x^{m-1-(m-1)}
        \\
        = {} & \hphantom{{} + {}}
        \sum_{i = 1}^{m}
        a_i c_{i,0} x^{m}
        +
        \sum_{\ell = 1}^{m-1}
        \sum_{i = 1}^{m}
        a_i c_{i,\ell} x^{m-\ell}
        \\
             &
        +
        \sum_{\ell = 0}^{m-2}
        \sum_{i = 1}^{m}
        b_i c_{i,(\ell+1)-1} x^{m-(\ell+1)}
        +
        \sum_{i = 1}^{m}
        b_i c_{i,m-1}
        \\
        = {} & \hphantom{{} + {}}
        \sum_{i = 1}^{m}
        a_i c_{i,0} x^{m}
        +
        \sum_{\ell = 1}^{m-1}
        \sum_{i = 1}^{m}
        a_i c_{i,\ell} x^{m-\ell}
        \\
             &
        +
        \sum_{\ell = 1}^{m-1}
        \sum_{i = 1}^{m}
        b_i c_{i,\ell-1} x^{m-\ell}
        +
        \sum_{i = 1}^{m}
        b_i c_{i,m-1}
        \\
        = {} &
        \sum_{i = 1}^{m} a_i c_{i,0}
        x^{m}
        +
        \sum_{\ell = 1}^{m-1}
        \sum_{i = 1}^{m}
        (a_i c_{i,\ell} + b_i c_{i,\ell-1})
        x^{m-\ell}
        +
        \sum_{i = 1}^{m} b_i c_{i,m-1}.
    \end{align*}
    记
    \begin{align*}
        d_k =
        \begin{dcases}
            \sum_{i = 1}^{m} a_i c_{i,0},
             & k = 0;     \\
            \sum_{i = 1}^{m}
            (a_i c_{i,k} + b_i c_{i,k-1}),
             & 0 < k < m; \\
            \sum_{i = 1}^{m} b_i c_{i,m-1},
             & k = m.
        \end{dcases}
    \end{align*}
    则 \(d_k\) 跟 \(A\), \(B\) 的元有关,
    且对任何数 \(x\), 有
    \begin{align*}
        \det {(xA + B)}
        = \sum_{k = 0}^{m} {d_k x^{m-k}}
        = d_0 x^m + d_1 x^{m-1} + \dots + d_m.
    \end{align*}
    最后, 不难算出
    \begin{align*}
         &
        d_0
        =
        \sum_{i = 1}^{m}
        a_i c_{i,0}
        =
        \sum_{i = 1}^{m}
        (-1)^{i+1} [A]_{i,1} \det {(A(i|1))}
        =
        \det {(A)},
        \\
         &
        d_m
        =
        \sum_{i = 1}^{m}
        b_i c_{i,m-1}
        =
        \sum_{i = 1}^{m}
        (-1)^{i+1} [B]_{i,1} \det {(B(i|1))}
        =
        \det {(B)}.
    \end{align*}
    所以, \(P(m)\) 是对的.
    由数学归纳法, 待证命题成立.
\end{example}

\begin{example}
    设 \(A\) 是 \(n\)~级阵.
    则存在跟 \(A\) 的元有关的数
    \(c_0\), \(c_1\), \(\dots\), \(c_n\),
    使对任何数 \(x\), 有
    \begin{align*}
        \det {(xI - A)}
        = \det {(xI + (-A))}
        = \sum_{k = 0}^{n} {c_k x^{n-k}}
        = c_0 x^n + c_1 x^{n-1} + \dots + c_n,
    \end{align*}
    其中,
    \(c_0 = \det {(I)} = 1\),
    且
    \(c_n = \det {(-A)} = (-1)^n \det {(A)}\).

    我们考虑 \(xI - A\) 的古伴
    \(\operatorname{adj} {(xI - A)}\).
    它的 \((i, j)\)-元
    \begin{align*}
        [\operatorname{adj} {(xI - A)}]_{i,j}
        = {} &
        (-1)^{j+i} \det {((xI - A)(j|i))}
        \\
        = {} &
        (-1)^{j+i} \det {(x I(j|i) + (-A)(j|i))}.
    \end{align*}
    则有跟 \(A\) 的元有关的数 \(d_{i,j,k}\)
    (\(k = 0\), \(1\), \(\dots\), \(n-1\)),
    使对任何数 \(x\), 有
    \begin{align*}
        \det {(x I(j|i) + (-A)(j|i))}
        = d_{i,j,0} x^{n-1}
        + d_{i,j,1} x^{n-2}
        + \dots + d_{i,j,n-1},
    \end{align*}
    其中,
    \(d_{i,j,0} = \det {(I(j|i))}\),
    且 \(d_{i,j,n-1} = \det {((-A)(j|i))}\).
    为方便, 记 \(b_{i,j,k} = (-1)^{j+i} d_{i,j,k}\).
    则有跟 \(A\) 的元有关的数 \(b_{i,j,k}\)
    (\(k = 0\), \(1\), \(\dots\), \(n-1\)),
    使对任何数 \(x\), 有
    \begin{align*}
        [\operatorname{adj} {(xI - A)}]_{i,j}
        = b_{i,j,0} x^{n-1}
        + b_{i,j,1} x^{n-2}
        + \dots + b_{i,j,n-1}.
    \end{align*}
    作 \(n\)~级阵 \(B_k\),
    使 \([B_k]_{i,j} = b_{i,j,k}\)
    (\(k = 0\), \(1\), \(\dots\), \(n-1\)).
    则 \(B_0\), \(B_1\), \(\dots\), \(B_{n-1}\)
    的元全跟 \(A\) 的元有关,
    且对任何数 \(x\), 有
    \begin{align*}
        \operatorname{adj} {(xI - A)}
        = x^{n-1} B_0 + x^{n-2} B_1 + \dots + B_{n-1}.
    \end{align*}
    因为
    \([B_{n-1}]_{i,j}
    = b_{i,j,n-1}
    = (-1)^{j+i} d_{i,j,n-1}
    = (-1)^{j+i} \det {((-A)(j|i))}\),
    我们有 \(B_{n-1} = \operatorname{adj} {(-A)}\).

    因为
    \begin{align*}
        (\operatorname{adj} {(xI - A)})\, (xI - A)
        = {} &
        \det {(xI - A)}\, I
        \\
        = {} &
        (c_0 x^n + c_1 x^{n-1} + \dots + c_n) I
        \\
        = {} &
        x^n (c_0 I) + x^{n-1} (c_1 I) + \dots + (c_n I),
    \end{align*}
    我们有
    \begin{align*}
             &
        (x^{n-1} B_0 + x^{n-2} B_1 + \dots + x B_{n-2} + B_{n-1})
        (xI - A)
        \\
        = {} &
        x^n (c_0 I) + x^{n-1} (c_1 I) + \dots + x (c_{n-1} I) + (c_n I).
    \end{align*}
    上式的左侧是
    \begin{align*}
             &
        (x^{n-1} B_0 + x^{n-2} B_1 + \dots + x B_{n-2} + B_{n-1})
        (xI)
        \\
             & \hphantom{(x^{n-1} B_0}
        +
        (x^{n-1} B_0 + x^{n-2} B_1 + \dots + x B_{n-2} + B_{n-1})
        (-A)
        \\
        = {} &
        x^n B_0 + x^{n-1} B_1 + \dots + x^2 B_{n-2} + x B_{n-1}
        \\
             &
        \hphantom{x^n B_0}
        - x^{n-1} (B_0 A) - x^{n-2} (B_1 A) - \dots
        - x (B_{n-2} A) - B_{n-1} A
        \\
        = {} &
        x^n B_0 + x^{n-1} (B_1 - B_0 A)
        + x^{n-2} (B_2 - B_1 A)
        \\
             &
        \qquad \qquad
        + \dots
        + x (B_{n-1} - B_{n-2} A)
        + (- B_{n-1} A).
    \end{align*}
    比较左侧与右侧, 有
    \begin{align*}
        B_0                 & = c_0 I,     \\
        B_1 - B_0 A         & = c_1 I,     \\
        B_2 - B_1 A         & = c_2 I,     \\
                            & \dots,       \\
        B_{n-1} - B_{n-2} A & = c_{n-1} I, \\
        - B_{n-1} A         & = c_n I.
    \end{align*}
    则
    \begin{align*}
        B_0 A^{n-1}               & = (c_0 I) A^{n-1}, \\
        (B_1 - B_0 A) A^{n-2}     & = (c_1 I) A^{n-2}, \\
        (B_2 - B_1 A) A^{n-3}     & = (c_2 I) A^{n-3}, \\
                                  & \dots,             \\
        (B_{n-1} - B_{n-2} A) A^0 & = (c_{n-1} I) A^0.
    \end{align*}
    则
    \begin{align*}
        B_0 A^{n-1}               & = c_0 A^{n-1}, \\
        B_1 A^{n-2} - B_0 A^{n-1} & = c_1 A^{n-2}, \\
        B_2 A^{n-3} - B_1 A^{n-2} & = c_2 A^{n-3}, \\
                                  & \dots,         \\
        B_{n-1} - B_{n-2} A       & = c_{n-1} I.
    \end{align*}
    则
    \begin{align*}
        B_{n-1} = c_0 A^{n-1} + c_1 A^{n-2} + \dots + c_{n-1} I.
    \end{align*}
    % 因为
    % \begin{align*}
    %     \operatorname{adj} {(0I - A)}
    %     = 0^{n-1} B_0 + 0^{n-2} B_1 + \dots + B_{n-1},
    % \end{align*}

    由 \(B_{n-1} = \operatorname{adj} {(-A)}\),
    知
    \begin{align*}
        \operatorname{adj} {(-A)}
        = c_0 A^{n-1} + c_1 A^{n-2} + \dots + c_{n-1} I.
    \end{align*}

    由 \(-B_{n-1} A = c_n I\), 知
    \begin{align*}
        0
        = {} &
        B_{n-1} A + c_n I
        \\
        = {} &
        (c_0 A^{n-1} + c_1 A^{n-2} + \dots + c_{n-1} I) A + c_n I
        \\
        = {} &
        c_0 A^{n} + c_1 A^{n-1} + \dots + c_{n-1} A + c_{n} I.
    \end{align*}

    % 我们知道,
    % \(\operatorname{adj} {(-A)}\, (-A) = \det {(-A)}\, I\).
    % 则
    % \begin{align*}
    %          &
    %     c_0 A^{n} + c_1 A^{n-1} + \dots + c_{n-1} A + c_{n} I
    %     \\
    %     = {} &
    %     (c_0 A^{n-1} + c_1 A^{n-2} + \dots + c_{n-1} I) A + c_{n} I
    %     \\
    %     = {} &
    %     \operatorname{adj} {(-A)}\, A + \det {(-A)}\, I
    %     \\
    %     = {} &
    %     {-} \operatorname{adj} {(-A)}\, (-A) + \det {(-A)}\, I
    %     \\
    %     = {} & 0.
    % \end{align*}

    最后, 由前面的例,
    \(1 \leq k \leq n\) 时,
    \begin{align*}
        c_k
        = {} &
        \sum_{1 \leq j_1 < \dots < j_k \leq n}
        \det {\left(
            (-A) \binom{j_1,\dots,j_k}{j_1,\dots,j_k}
            \right)}
        \\
        = {} &
        \sum_{1 \leq j_1 < \dots < j_k \leq n}
        (-1)^k
        \det {\left(
            A \binom{j_1,\dots,j_k}{j_1,\dots,j_k}
            \right)}
        \\
        = {} &
        (-1)^k
        \sum_{1 \leq j_1 < \dots < j_k \leq n}
        \det {\left(
            A \binom{j_1,\dots,j_k}{j_1,\dots,j_k}
            \right)};
    \end{align*}
    特别地, \(c_n = (-1)^n \det {(A)}\).
\end{example}

综上, 我们有:

\begin{theorem}
    设 \(A\) 是 \(n\)~级阵.

    (1)
    存在跟 \(A\) 的元有关的数
    \(c_0\), \(c_1\), \(\dots\), \(c_n\),
    使对任何数 \(x\), 有
    \begin{align*}
        \det {(xI - A)} = c_0 x^n + c_1 x^{n-1} + \dots + c_n,
    \end{align*}
    其中, \(c_0 = 1\),
    且 \(c_n = (-1)^n \det {(A)}\).
    更具体地,
    \(1 \leq k \leq n\) 时,
    \begin{align*}
        c_k
        = {} &
        (-1)^k
        \sum_{1 \leq j_1 < \dots < j_k \leq n}
        \det {\left(
            A \binom{j_1,\dots,j_k}{j_1,\dots,j_k}
            \right)}.
    \end{align*}

    (2)
    用 \(A\) 的一些非负整数次方的数乘的和,
    我们可如此表示 \(-A\) 的古伴:
    \begin{align*}
        \operatorname{adj} {(-A)}
        = c_0 A^{n-1} + c_1 A^{n-2} + \dots + c_{n-1} I.
    \end{align*}
    则 \(A\) 的古伴
    \begin{align*}
        \operatorname{adj} {(A)}
        = (-1)^{n-1} \operatorname{adj} {(-A)}
        = (-1)^{n-1} (c_0 A^{n-1} + c_1 A^{n-2} + \dots + c_{n-1} I).
    \end{align*}

    (3)
    (Cayley--Hamilton \pinjino{keili--hamerten} 定理)
    \begin{align*}
        c_0 A^{n} + c_1 A^{n-1} + \dots + c_{n} I = 0.
    \end{align*}
\end{theorem}

\begin{example}
    我们验证 \(n = 2\) 时的情形.
    为方便, 设
    \(A =
    \begin{bmatrix}
        a & b \\
        c & d \\
    \end{bmatrix}\).
    则
    \(xI - A =
    \begin{bmatrix}
        x - a & -b    \\
        -c    & x - d \\
    \end{bmatrix}\).
    直接计算, 有
    \begin{align*}
        \det {(xI - A)}
        = (x - a)(x - d) - (-c)(-b)
        = 1x^2 + (-(a + d))x + (ad - cb).
    \end{align*}
    注意, \((-1)^2 \det {(A)} = ad - cb\),
    且
    \begin{align*}
        (-1)^1
        \sum_{1 \leq j_1 \leq 2}
        \det {\left(
            A \binom{j_1}{j_1}
            \right)}
        = {} &
        (-1) \left(
        \det {\left(
            A \binom{1}{1}
            \right)}
        +
        \det {\left(
            A \binom{2}{2}
            \right)}
        \right)
        \\
        = {} &
        {-} (a + d).
    \end{align*}
    则 (1) 被验证.

    因为
    \(-A =
    \begin{bmatrix}
        -a & -b \\
        -c & -d \\
    \end{bmatrix}\),
    我们有
    \begin{align*}
        \operatorname{adj} {(-A)}
        =
        \begin{bmatrix}
            -d & b  \\
            c  & -a \\
        \end{bmatrix}
        =
        \begin{bmatrix}
            a - (a+d) & b         \\
            c         & d - (a+d) \\
        \end{bmatrix}
        =
        1A + (-(a+d))I.
    \end{align*}
    类似地, 可验证
    \(\operatorname{adj} {(A)}
    = (-1)^{2-1} \operatorname{adj} {(-A)}
    = -(1A + (-(a+d))I)\);
    请允许我留它为您的习题.
    则 (2) 被验证.

    最后,
    \begin{align*}
             &
        1 A^2 + (-(a + d)) A + (ad - cb) I
        \\
        = {} &
        1
        \begin{bmatrix}
            a & b \\
            c & d \\
        \end{bmatrix}
        \begin{bmatrix}
            a & b \\
            c & d \\
        \end{bmatrix}
        + (-(a + d))
        \begin{bmatrix}
            a & b \\
            c & d \\
        \end{bmatrix}
        + (ad - cb)
        \begin{bmatrix}
            1 & 0 \\
            0 & 1 \\
        \end{bmatrix}
        \\
        = {} &
        \begin{bmatrix}
            a^2 + bc & b(a + d) \\
            c(a + d) & bc + d^2 \\
        \end{bmatrix}
        +
        \begin{bmatrix}
            -a^2 - ad & -b(a + d) \\
            -c(a + d) & -ad - d^2 \\
        \end{bmatrix}
        +
        \begin{bmatrix}
            ad - bc & 0       \\
            0       & ad - bc \\
        \end{bmatrix}
        \\
        = {} &
        \begin{bmatrix}
            0 & 0 \\
            0 & 0 \\
        \end{bmatrix}
        \\
        = {} &
        0.
    \end{align*}
    则 (3) 被验证.
\end{example}

% \begin{example}
%     设 \(A\) 是 \(2m\)~级反称阵,
%     且 \(\det {(A)} = 0\).
%     则 \(\operatorname{adj} {(A)} = 0\).

%     注意, \(A^{\mathrm{T}} = -A\), 则
%     \begin{align*}
%         (\operatorname{adj} {(A)})^{\mathrm{T}}
%         = \operatorname{adj} {(A^{\mathrm{T}})}
%         = \operatorname{adj} {(-A)}
%         = (-1)^{2m-1} \operatorname{adj} {(A)}
%         = -\operatorname{adj} {(A)}.
%     \end{align*}
%     注意, \(A(i|i)\) 是奇数级反称阵,
%     则它的行列式是 \(0\).
%     则 \([\operatorname{adj} {(A)}]_{i,i} = 0\).
%     则 \(\operatorname{adj} {(A)}\) 也是反称阵.

%     既然 \(A\) 的级不低于 \(2\),
%     且 \(\det {(A)} = 0\),
%     则 \(\operatorname{adj} {(A)}\) 的%
%     每个 \(2\)~级子阵的行列式是 \(0\)
%     (见 ``古伴的性质 (2)'').
%     那么, 对 \(1 \leq i < j \leq 2m\),
%     \begin{align*}
%         0
%         = \det {\left(
%             (\operatorname{adj} {(A)})
%             \binom{i,j}{i,j}
%             \right)}
%         = \det {
%             \begin{bmatrix}
%                 0                                 & [\operatorname{adj} {(A)}]_{i,j} \\
%                 -[\operatorname{adj} {(A)}]_{i,j} & 0                                \\
%             \end{bmatrix}
%         }
%         = ([\operatorname{adj} {(A)}]_{i,j})^2.
%     \end{align*}
%     则 \([\operatorname{adj} {(A)}]_{i,j} = 0\).
%     则 \([\operatorname{adj} {(A)}]_{j,i} = 0\).
%     则 \(\operatorname{adj} {(A)} = 0\).
% \end{example}

\begin{example}
    设 \(A\) 是 \(n\)~级阵.
    设 \(i\) 是不超过 \(n\) 的正整数.
    按行~\(i\) 展开, 有
    \begin{align*}
        \det {(A)}
        = [A]_{i,i} \det {(A(i|i))}
        + \sum_{\substack{1 \leq \ell \leq n \\ \ell \neq i}}
        {(-1)^{i+\ell}\, [A]_{i,\ell} \det {(A(i|\ell))}}.
    \end{align*}
    注意, \(A\) 的列~\(i\) 对应
    \(A(i|\ell)\) 的列~\(i - \rho(i, \ell)\).
    则
    \begin{align*}
        \det {(A(i|\ell))}
        = \sum_{\substack{1 \leq k \leq n \\ k \neq i}}
        {(-1)^{k - \rho(k, i) + i - \rho(i, \ell)}\,
        [A]_{k,i} \det {(A({i,k}|{\ell,i}))}}.
    \end{align*}
    所以 (注意, 对整数 \(p\), \(q\),
    有 \((-1)^{p+q} = (-1)^{p-q}\))
    \begin{align*}
             &
        \det {(A)}
        \\
        = {} &
        [A]_{i,i} \det {(A(i|i))}
        + \sum_{\substack{
        1 \leq k, \ell \leq n \\
            k, \ell \neq i}}
        {(-1)^{k+\ell-\rho(k,i)+\rho(i,\ell)}\,
        [A]_{k,i} [A]_{i,\ell} \det {(A({i,k}|{i,\ell}))}}.
    \end{align*}

    设 \(A\) 是反称阵.
    则 \([A]_{i,i} = 0\),
    且 \([A]_{k,i} = -[A]_{i,k}\).
    注意, \(1 - \rho(k, i) = \rho(i, k)\).
    则
    \begin{align*}
        \det {(A)}
        =
        \sum_{\substack{
        1 \leq k, \ell \leq n \\
            k, \ell \neq i}}
        {(-1)^{k+\ell+\rho(i,k)+\rho(i,\ell)}\,
        [A]_{i,k} [A]_{i,\ell} \det {(A({i,k}|{i,\ell}))}}.
    \end{align*}
    则
    \begin{align*}
        %  &
        \det {(A)}
        % \\
        = {} &
        \hphantom{{} + {}}
        \sum_{\substack{
        1 \leq k, \ell \leq n  \\
        k, \ell \neq i         \\
            k = \ell}}
        {(-1)^{k+\ell+\rho(i,k)+\rho(i,\ell)}\,
        [A]_{i,k} [A]_{i,\ell} \det {(A({i,k}|{i,\ell}))}}
        \\
             &
        +
        \sum_{\substack{
        1 \leq k, \ell \leq n  \\
        k, \ell \neq i         \\
            k < \ell}}
        {(-1)^{k+\ell+\rho(i,k)+\rho(i,\ell)}\,
        [A]_{i,k} [A]_{i,\ell} \det {(A({i,k}|{i,\ell}))}}
        \\
             &
        +
        \sum_{\substack{
        1 \leq k, \ell \leq n  \\
        k, \ell \neq i         \\
            k > \ell}}
        {(-1)^{k+\ell+\rho(i,k)+\rho(i,\ell)}\,
        [A]_{i,k} [A]_{i,\ell} \det {(A({i,k}|{i,\ell}))}}
        \\
        = {} &
        \hphantom{{} + {}}
        \sum_{\substack{
        1 \leq k \leq n        \\
            k \neq i}}
        { [A]_{i,k}^2 \det {(A({i,k}|{i,k}))}}
        \\
             &
        +
        \sum_{\substack{
        1 \leq k, \ell \leq n  \\
        k, \ell \neq i         \\
            k < \ell}}
        {(-1)^{k+\ell+\rho(i,k)+\rho(i,\ell)}\,
        [A]_{i,k} [A]_{i,\ell} \det {(A({i,k}|{i,\ell}))}}
        \\
             &
        +
        \sum_{\substack{
        1 \leq k, \ell \leq n  \\
        k, \ell \neq i         \\
            k < \ell}}
        {(-1)^{\ell+k+\rho(i,\ell)+\rho(i,k)}\,
        [A]_{i,\ell} [A]_{i,k} \det {(A({i,\ell}|{i,k}))}}
        \\
        = {} &
        \hphantom{{} + {}}
        \sum_{\substack{
        1 \leq k \leq n        \\
            k \neq i}}
        { [A]_{i,k}^2 \det {(A({i,k}|{i,k}))}}
        \\
             &
        +
        \sum_{\substack{
        1 \leq k < \ell \leq n \\
            k, \ell \neq i}}
        {(-1)^{k+\ell+\rho(i,k)+\rho(i,\ell)}\,
        [A]_{i,k} [A]_{i,\ell} \det {(A({i,k}|{i,\ell}))}}
        \\
             &
        +
        \sum_{\substack{
        1 \leq k < \ell \leq n \\
            k, \ell \neq i}}
        {(-1)^{k+\ell+\rho(i,k)+\rho(i,\ell)}\,
        [A]_{i,k} [A]_{i,\ell}
        \det {(A({i,\ell}|{i,k}))}}.
    \end{align*}
    注意, \((A({i,\ell}|{i,k}))^{\mathrm{T}}
    = (A^{\mathrm{T}}) ({i,k}|{i,\ell})
    = (-A) ({i,k}|{i,\ell})
    = - (A({i,k}|{i,\ell}))\).
    则
    \begin{align*}
        \det {(A({i,\ell}|{i,k}))}
        = {} &
        \det {(A({i,\ell}|{i,k}))^{\mathrm{T}})}
        \\
        = {} &
        (-1)^{n-2} \det {(A({i,k}|{i,\ell}))}
        \\
        = {} &
        (-1)^{n} \det {(A({i,k}|{i,\ell}))}.
    \end{align*}
    则
    \begin{align*}
             &
        \det {(A)}
        \\
        = {} &
        \hphantom{{} + {}}
        \sum_{\substack{
        1 \leq k \leq n        \\
            k \neq i}}
        { [A]_{i,k}^2 \det {(A({i,k}|{i,k}))}}
        \\
             &
        +
        (1 + (-1)^n)
        \sum_{\substack{
        1 \leq k < \ell \leq n \\
            k, \ell \neq i}}
        {(-1)^{k+\ell+\rho(i,k)+\rho(i,\ell)}\,
        [A]_{i,k} [A]_{i,\ell} \det {(A({i,k}|{i,\ell}))}}.
    \end{align*}
    回想, 对反称阵 \(A\), 有
    \(\det {(A)} = (\operatorname{pf} {(A)})^2\).
    回想,
    \begin{align*}
        \operatorname{pf} {(A)}
        =
        \sum_{\substack{1 \leq j \leq n \\j \neq i}}
        {
        (-1)^{i-1+j+\rho(i,j)}\, [A]_{i,j}
        \operatorname{pf} {(A({i,j}|{i,j}))}}.
    \end{align*}
    回想,
    \begin{align*}
        (x_1 + x_2 + \dots + x_n)^2
        = (x_1^2 + x_2^2 + \dots + x_n^2)
        + 2 \sum_{1 \leq k < \ell \leq n} {x_k x_\ell}.
    \end{align*}
    则
    \begin{align*}
             &
        (\operatorname{pf} {(A)})^2
        \\
        = {} &
        \hphantom{{} + {}}
        \sum_{\substack{
        1 \leq k \leq n        \\
            k \neq i}}
        {[A]_{i,k}^2
        (\operatorname{pf} {(A({i,k}|{i,k}))})^2}
        \\
             &
        +
        2
        \sum_{\substack{
        1 \leq k < \ell \leq n \\
            k, \ell \neq i}}
        {
        (-1)^{k+\ell+\rho(i,k)+\rho(i,\ell)}\,
        [A]_{i,k} [A]_{i,\ell}
        \operatorname{pf} {(A({i,k}|{i,k}))}
        \operatorname{pf} {(A({i,\ell}|{i,\ell}))}
        }.
    \end{align*}
    注意, \(\det {(A({i,k}|{i,k}))}
    = (\operatorname{pf} {(A({i,k}|{i,k}))})^2\).
    则
    \begin{align*}
             &
        (1 + (-1)^n)
        \sum_{\substack{
        1 \leq k < \ell \leq n \\
            k, \ell \neq i}}
        {(-1)^{k+\ell+\rho(i,k)+\rho(i,\ell)}\,
        [A]_{i,k} [A]_{i,\ell} \det {(A({i,k}|{i,\ell}))}}
        \\
        = {} &
        2
        \sum_{\substack{
        1 \leq k < \ell \leq n \\
            k, \ell \neq i}}
        {
        (-1)^{k+\ell+\rho(i,k)+\rho(i,\ell)}\,
        [A]_{i,k} [A]_{i,\ell}
        \operatorname{pf} {(A({i,k}|{i,k}))}
        \operatorname{pf} {(A({i,\ell}|{i,\ell}))}
        }.
    \end{align*}

    以下, 设 \(n\) 是偶数.
    则
    \begin{align*}
             &
        \sum_{\substack{
        1 \leq k < \ell \leq n \\
            k, \ell \neq i}}
        {(-1)^{k+\ell+\rho(i,k)+\rho(i,\ell)}\,
        [A]_{i,k} [A]_{i,\ell} \det {(A({i,k}|{i,\ell}))}}
        \\
        = {} &
        \sum_{\substack{
        1 \leq k < \ell \leq n \\
            k, \ell \neq i}}
        {
        (-1)^{k+\ell+\rho(i,k)+\rho(i,\ell)}\,
        [A]_{i,k} [A]_{i,\ell}
        \operatorname{pf} {(A({i,k}|{i,k}))}
        \operatorname{pf} {(A({i,\ell}|{i,\ell}))}
        }.
    \end{align*}

    设 \(1 \leq u < v \leq n\),
    \(u \neq i\), 且 \(v \neq i\).
    作 \(n\) 级反称阵 \(B\),
    其中,
    \begin{align*}
        [B]_{p,q}
        = \begin{cases}
              1,
               & \text{\(p = i\), 且 \(q = u\),
              或 \(p = i\), 且 \(q = v\)};             \\
              -1,
               & \text{\(p = u\), 且 \(q = i\),
              或 \(p = v\), 且 \(q = i\)};             \\
              0,
               & \text{\(p = i\), 且 \(q \neq u, v\),
              或 \(q = i\), 且 \(p \neq u, v\)};       \\
              [A]_{p,q},
               & \text{别的情形}.
          \end{cases}
    \end{align*}
    注意, \(A(i|i) = B(i|i)\).
    则
    \(A({i,k}|{i,k}) = B({i,k}|{i,k})\),
    \(A({i,\ell}|{i,\ell}) = B({i,\ell}|{i,\ell})\),
    且
    \(A({i,k}|{i,\ell}) = B({i,k}|{i,\ell})\).
    则, 由
    \begin{align*}
             &
        \sum_{\substack{
        1 \leq k < \ell \leq n \\
            k, \ell \neq i}}
        {(-1)^{k+\ell+\rho(i,k)+\rho(i,\ell)}\,
        [B]_{i,k} [B]_{i,\ell} \det {(B({i,k}|{i,\ell}))}}
        \\
        = {} &
        \sum_{\substack{
        1 \leq k < \ell \leq n \\
            k, \ell \neq i}}
        {
        (-1)^{k+\ell+\rho(i,k)+\rho(i,\ell)}\,
        [B]_{i,k} [B]_{i,\ell}
        \operatorname{pf} {(B({i,k}|{i,k}))}
        \operatorname{pf} {(B({i,\ell}|{i,\ell}))}
        },
    \end{align*}
    我们有
    \begin{align*}
             &
        \sum_{\substack{
        1 \leq k < \ell \leq n \\
            k, \ell \neq i}}
        {(-1)^{k+\ell+\rho(i,k)+\rho(i,\ell)}\,
        [B]_{i,k} [B]_{i,\ell} \det {(A({i,k}|{i,\ell}))}}
        \\
        = {} &
        \sum_{\substack{
        1 \leq k < \ell \leq n \\
            k, \ell \neq i}}
        {
        (-1)^{k+\ell+\rho(i,k)+\rho(i,\ell)}\,
        [B]_{i,k} [B]_{i,\ell}
        \operatorname{pf} {(A({i,k}|{i,k}))}
        \operatorname{pf} {(A({i,\ell}|{i,\ell}))}
        }.
    \end{align*}
    注意, 若 \(k \neq u\), 或 \(k \neq v\),
    或 \(\ell \neq u\), 或 \(\ell \neq v\),
    则 \([B]_{i,k} [B]_{i,\ell} = 0\).
    所以
    \begin{align*}
             &
        (-1)^{u+v+\rho(i,u)+\rho(i,v)}\,
        1 \cdot 1 \cdot
        \det {(A({i,u}|{i,v}))}
        \\
        = {} &
        (-1)^{u+v+\rho(i,u)+\rho(i,v)}\,
        1 \cdot 1 \cdot
        \operatorname{pf} {(A({i,u}|{i,u}))}
        \operatorname{pf} {(A({i,v}|{i,v}))},
    \end{align*}
    即
    \begin{align*}
        \det {(A({i,u}|{i,v}))}
        = \operatorname{pf} {(A({i,u}|{i,u}))}
        \operatorname{pf} {(A({i,v}|{i,v}))}.
    \end{align*}

    % 设 \(1 \leq u < v \leq n\),
    % \(u \neq i\), 且 \(v \neq i\).
    % 作 \(n\) 级反称阵 \(B\),
    % 其中,
    % \begin{align*}
    %     [B]_{p,q}
    %     = \begin{cases}
    %           x,         & \text{\(p = i\), 且 \(q = u\)}; \\
    %           x,         & \text{\(p = i\), 且 \(q = v\)}; \\
    %           -x,        & \text{\(p = u\), 且 \(q = i\)}; \\
    %           -x,        & \text{\(p = v\), 且 \(q = i\)}; \\
    %           [A]_{p,q}, & \text{别的情形}.
    %       \end{cases}
    % \end{align*}
    % 注意, \(A(i|i) = B(i|i)\).
    % 则
    % \(A({i,k}|{i,k}) = B({i,k}|{i,k})\),
    % \(A({i,\ell}|{i,\ell}) = B({i,\ell}|{i,\ell})\),
    % 且
    % \(A({i,k}|{i,\ell}) = B({i,k}|{i,\ell})\).
    % 则, 由
    % \begin{align*}
    %          &
    %     \sum_{\substack{
    %     1 \leq k < \ell \leq n \\
    %         k, \ell \neq i}}
    %     {(-1)^{k+\ell+\rho(i,k)+\rho(i,\ell)}\,
    %         [B]_{i,k} [B]_{i,\ell} \det {(B({i,k}|{i,\ell}))}}
    %     \\
    %     = {} &
    %     \sum_{\substack{
    %     1 \leq k < \ell \leq n \\
    %         k, \ell \neq i}}
    %     {
    %     (-1)^{k+\ell+\rho(i,k)+\rho(i,\ell)}\,
    %         [B]_{i,k} [B]_{i,\ell}
    %     \operatorname{pf} {(B({i,k}|{i,k}))}
    %     \operatorname{pf} {(B({i,\ell}|{i,\ell}))}
    %     },
    % \end{align*}
    % 我们有
    % \begin{align*}
    %          &
    %     \sum_{\substack{
    %     1 \leq k < \ell \leq n \\
    %         k, \ell \neq i}}
    %     {(-1)^{k+\ell+\rho(i,k)+\rho(i,\ell)}\,
    %         [B]_{i,k} [B]_{i,\ell} \det {(A({i,k}|{i,\ell}))}}
    %     \\
    %     = {} &
    %     \sum_{\substack{
    %     1 \leq k < \ell \leq n \\
    %         k, \ell \neq i}}
    %     {
    %     (-1)^{k+\ell+\rho(i,k)+\rho(i,\ell)}\,
    %         [B]_{i,k} [B]_{i,\ell}
    %     \operatorname{pf} {(A({i,k}|{i,k}))}
    %     \operatorname{pf} {(A({i,\ell}|{i,\ell}))}
    %     }.
    % \end{align*}
    % 这是关于 \(x\) 的 \({\leq} 2\)~次式的等式,
    % 因为可能含 \(x\) 的 \([B]_{i,k} [B]_{i,\ell}\)
    % 的次不超过 \(2\).
    % 左侧的 \({\leq} 2\)~次式的 \(x^2\)~项的系数是
    % \((-1)^{u+v+\rho(i,u)+\rho(i,v)}\,
    % \det {(A({i,u}|{i,v}))}\);
    % 右侧的 \({\leq} 2\)~次式的 \(x^2\)~项的系数是
    % \((-1)^{u+v+\rho(i,u)+\rho(i,v)}\,
    % \operatorname{pf} {(A({i,u}|{i,u}))}
    % \operatorname{pf} {(A({i,v}|{i,v}))}\).
    % 则它们应相等.
    % 则 \(1 \leq u < v \leq n\),
    % \(u \neq i\), 且 \(v \neq i\) 时,
    % \begin{align*}
    %     \det {(A({i,u}|{i,v}))}
    %     = \operatorname{pf} {(A({i,u}|{i,u}))}
    %     \operatorname{pf} {(A({i,v}|{i,v}))}.
    % \end{align*}

    回想,
    \begin{align*}
        \det {(A({i,u}|{i,v}))}
        = (-1)^n \det {(A({i,v}|{i,u}))}
        = \det {(A({i,v}|{i,u}))}.
    \end{align*}
    则 \(1 \leq v < u \leq n\),
    \(u \neq i\), 且 \(v \neq i\) 时,
    \begin{align*}
        \det {(A({i,u}|{i,v}))}
        = {} &
        \det {(A({i,v}|{i,u}))}
        \\
        = {} &
        \operatorname{pf} {(A({i,v}|{i,v}))}
        \operatorname{pf} {(A({i,u}|{i,u}))}
        \\
        = {} &
        \operatorname{pf} {(A({i,u}|{i,u}))}
        \operatorname{pf} {(A({i,v}|{i,v}))}.
    \end{align*}
    回想, \(\det {(A({i,u}|{i,u}))}
    = (\operatorname{pf} {(A({i,u}|{i,u}))})^2\).
    于是, 对不等于 \(i\) 的,
    且不超过 \(n\) 的正整数 \(u\), \(v\),
    \begin{align*}
        \det {(A({i,u}|{i,v}))}
        = \operatorname{pf} {(A({i,u}|{i,u}))}
        \operatorname{pf} {(A({i,v}|{i,v}))}.
    \end{align*}
    所以, 若 \(j \neq i\),
    \begin{align*}
             &
        \det {(A(i|j))}
        \\
        = {} &
        \sum_{\substack{1 \leq k \leq n \\ k \neq i}}
        {(-1)^{k - \rho(k, i) + i - \rho(i, j)}\,
        [A]_{k,i} \det {(A({i,k}|{i,j}))}}
        \\
        = {} &
        \sum_{\substack{1 \leq k \leq n \\ k \neq i}}
        {(-1)^{k - \rho(k, i) + i - \rho(i, j)}\,
        [A]_{k,i}
        \operatorname{pf} {(A({i,k}|{i,k}))}
        \operatorname{pf} {(A({i,j}|{i,j}))}}
        \\
        = {} &
        \sum_{\substack{1 \leq k \leq n \\ k \neq i}}
        {(-1)^{i + k + 1 - \rho(i, k)} (-1)^{1 - \rho(i, j)}\,
        [A]_{i,k}
        \operatorname{pf} {(A({i,k}|{i,k}))}
        \operatorname{pf} {(A({i,j}|{i,j}))}}
        \\
        = {} &
        (-1)^{\rho(j, i)}
        \operatorname{pf} {(A({i,j}|{i,j}))}
        \sum_{\substack{1 \leq k \leq n \\ k \neq i}}
        {(-1)^{i - 1 + k + \rho(i, k)}\,
        [A]_{i,k}
        \operatorname{pf} {(A({i,k}|{i,k}))}}
        \\
        = {} &
        (-1)^{\rho(j, i)}
        \operatorname{pf} {(A({i,j}|{i,j}))}
        \operatorname{pf} {(A)}.
    \end{align*}
\end{example}

综上, 我们有:

\begin{theorem}
    设 \(A\) 是 \(2m\)~级反称阵.
    设 \(i\) 是不超过 \(2m\) 的正整数.
    设不超过 \(2m\) 的正整数 \(k\), \(\ell\) 不等于 \(i\).
    则
    \begin{align*}
        \det {(A({i,k}|{i,\ell}))}
        = \operatorname{pf} {(A({i,k}|{i,k}))}
        \operatorname{pf} {(A({i,\ell}|{i,\ell}))},
    \end{align*}
    且
    \begin{align*}
        \det {(A(i|\ell))}
        = (-1)^{\rho(\ell, i)}
        \operatorname{pf} {(A({i,\ell}|{i,\ell}))}
        \operatorname{pf} {(A)}.
    \end{align*}
    则
    \begin{align*}
        [\operatorname{adj} {(A)}]_{i,j}
        = {} &
        (-1)^{j+i} \det {(A(j|i))}
        \\
        = {} &
        \begin{cases}
            (-1)^{\rho(i, j)}
            (-1)^{i+j}
            \operatorname{pf} {(A({i,j}|{i,j}))}
            \operatorname{pf} {(A)},
             & i \neq j; \\
            0,
             & i = j.
        \end{cases}
    \end{align*}
\end{theorem}

\begin{example}
    设 \(A\) 是 \(2m-1\)~级反称阵.
    作 \(2m\)~级反称阵 \(B\), 其中,
    \begin{align*}
        [B]_{i,j} =
        \begin{cases}
            0,         & \text{\(i = 2m\), 或 \(j = 2m\)}; \\
            [A]_{i,j}, & \text{别的情形}.
        \end{cases}
    \end{align*}
    则 \(B({2m}|{2m}) = A\).
    设 \(i\), \(j\) 是不超过 \(2m-1\) 的正整数.
    则
    \begin{align*}
        \det {(A(j|i))}
        = {} &
        \det {(B({2m,j}|{2m,i}))}
        \\
        = {} &
        \operatorname{pf} {(B({2m,j}|{2m,j}))}
        \operatorname{pf} {(B({2m,i}|{2m,i}))}
        \\
        = {} &
        \operatorname{pf} {(A(j|j))}
        \operatorname{pf} {(A(i|i))}.
    \end{align*}
    则
    \begin{align*}
        [\operatorname{adj} {(A)}]_{i,j}
        = {} &
        (-1)^{j+i} \det {(A(j|i))}
        \\
        = {} &
        (-1)^i \operatorname{pf} {(A(i|i))}
        \cdot
        (-1)^j \operatorname{pf} {(A(j|j))}.
    \end{align*}
    于是, 若 \((2m-1) \times 1\)~阵 \(X\) 适合
    \([X]_{i,1}
    = (-1)^i \operatorname{pf} {(A(i|i))}\),
    则 \(\operatorname{adj} {(A)} = X X^{\mathrm{T}}\).
\end{example}

\begin{example}
    设 \(A\) 是 \(2m\)~级反称阵.
    它的古伴 \(\operatorname{adj} {(A)}\)
    也是反称阵,
    故 \(\operatorname{adj} {(A)}\) 的 pfaffian 是有意义的.
    则
    \begin{align*}
        \operatorname{pf} {(A^{\mathrm{T}} \operatorname{adj} {(A)}\, A)}
        = \operatorname{pf} {(\operatorname{adj} {(A)})} \det {(A)}.
    \end{align*}
    注意,
    \begin{align*}
        A^{\mathrm{T}} \operatorname{adj} {(A)}\, A
        = (-A) (\det {(A)}\, I)
        = (-\det {(A)}) A,
    \end{align*}
    故
    \(\operatorname{pf} {(A^{\mathrm{T}} \operatorname{adj} {(A)}\, A)}
    = (-\det {(A)})^m \operatorname{pf} {(A)}\).
    于是, 若 \(\det {(A)} \neq 0\),
    \begin{align*}
        \operatorname{pf} {(\operatorname{adj} {(A)})}
        = (-1)^m\, (\det {(A)})^{m-1} \operatorname{pf} {(A)}
        = (-1)^m\, (\operatorname{pf} {(A)})^{2m-1}.
    \end{align*}
    若 \(\det {(A)} = 0\),
    则 \(\operatorname{pf} {(A)} = 0\),
    且由前面的结果, \(\operatorname{adj} {(A)} = 0\).
    则
    \(\operatorname{pf} {(\operatorname{adj} {(A)})} = 0\).
    故上式仍是对的.
\end{example}

\begin{example}
    设一元运算 \(\mathrm{D}\) 适合,
    对任何 \(a\), \(b\), 必
    \begin{align*}
        \mathrm{D} (a + b)     & =
        \mathrm{D} (a) + \mathrm{D} (b), \\
        \mathrm{D} (a \cdot b) & =
        \mathrm{D} (a) \cdot b + a \cdot \mathrm{D} (b).
    \end{align*}

    由此, \(\mathrm{D} (1)
    = \mathrm{D} (1 \cdot 1)
    = \mathrm{D} (1) \cdot 1 + 1 \cdot \mathrm{D} (1)
    = \mathrm{D} (1) + \mathrm{D} (1)\).
    则 \(\mathrm{D} (1) = 0\).

    注意, \(\mathrm{D} (0)
    = \mathrm{D} (0 + 0)
    = \mathrm{D} (0) + \mathrm{D} (0)\).
    则 \(\mathrm{D} (0) = 0\).
    则 \(\mathrm{D} (0 \cdot x)
    = \mathrm{D} (0) = 0 = 0 \cdot \mathrm{D} (x)\).

    注意, \(\mathrm{D} (0)
    = \mathrm{D} (x + (-x))
    = \mathrm{D} (x) + \mathrm{D} (-x)\).
    则 \(\mathrm{D} (-x) = -\mathrm{D} (x)\).
    则 \(\mathrm{D} (\pm x) = \pm \mathrm{D} (x)\).
    则 \(\mathrm{D} (x \pm y)
    = \mathrm{D} (x) + \mathrm{D} (\pm y)
    = \mathrm{D} (x) \pm \mathrm{D} (y)\).
    于是, 若 \(s_1\), \(s_2\), \(\dots\), \(s_n\)
    是 \(1\), \(0\), 或 \(-1\),
    则
    \begin{align*}
        \mathrm{D} \left(
        \sum_{\ell = 1}^{n} {s_\ell x_\ell}
        \right)
        = \sum_{\ell = 1}^{n} {s_\ell \mathrm{D} (x_\ell)}.
    \end{align*}

    设 \(A\) 是 \(n\)~级阵.
    我们研究, \(\mathrm{D} (\det {(A)})\) 会是何.

    设 \(n = 1\).
    则 \(\det {(A)} = [A]_{1,1}\).
    则 \(\mathrm{D} (\det {(A)})
    = \mathrm{D} ([A]_{1,1})
    = \det {[\mathrm{D} ([A]_{1,1})]}\).

    设 \(n = 2\).
    则 \(\det {(A)} = [A]_{1,1} [A]_{2,2} - [A]_{2,1} [A]_{1,2}\).
    则
    \begin{align*}
             &
        \mathrm{D} (\det {(A)})
        \\
        = {} &
        \mathrm{D} ([A]_{1,1} [A]_{2,2} - [A]_{2,1} [A]_{1,2})
        \\
        = {} &
        \mathrm{D} ([A]_{1,1} [A]_{2,2})
        - \mathrm{D} ([A]_{2,1} [A]_{1,2})
        \\
        = {} &
        (\mathrm{D} ([A]_{1,1}) [A]_{2,2} + [A]_{1,1} \mathrm{D} ([A]_{2,2}))
        -
        (\mathrm{D} ([A]_{2,1}) [A]_{1,2} + [A]_{2,1} \mathrm{D} ([A]_{1,2}))
        \\
        = {} &
        (\mathrm{D} ([A]_{1,1}) [A]_{2,2} - [A]_{2,1} \mathrm{D} ([A]_{1,2}))
        +
        ([A]_{1,1} \mathrm{D} ([A]_{2,2}) - \mathrm{D} ([A]_{2,1}) [A]_{1,2})
        \\
        = {} &
        \det {\begin{bmatrix}
                      \mathrm{D} ([A]_{1,1}) & [A]_{1,2} \\
                      \mathrm{D} ([A]_{2,1}) & [A]_{2,2} \\
                  \end{bmatrix}}
        +
        \det {\begin{bmatrix}
                      [A]_{1,1} & \mathrm{D} ([A]_{1,2}) \\
                      [A]_{2,1} & \mathrm{D} ([A]_{2,2}) \\
                  \end{bmatrix}}.
    \end{align*}

    设 \(n = 3\).
    则
    \begin{align*}
        \det {(A)}
        = {} &
        \hphantom{{} + {}}
        [A]_{1,1}
        \det {\begin{bmatrix}
                      [A]_{2,2} & [A]_{2,3} \\
                      [A]_{3,2} & [A]_{3,3} \\
                  \end{bmatrix}}
        - [A]_{2,1}
        \det {\begin{bmatrix}
                      [A]_{1,2} & [A]_{1,3} \\
                      [A]_{3,2} & [A]_{3,3} \\
                  \end{bmatrix}}
        \\
             &
        + [A]_{3,1}
        \det {\begin{bmatrix}
                      [A]_{1,2} & [A]_{1,3} \\
                      [A]_{2,2} & [A]_{2,3} \\
                  \end{bmatrix}}.
    \end{align*}
    则
    \begin{align*}
             &
        \mathrm{D} (\det {(A)})
        \\
        = {} &
        \hphantom{{} + {}}
        \mathrm{D} ([A]_{1,1})
        \det {\begin{bmatrix}
                      [A]_{2,2} & [A]_{2,3} \\
                      [A]_{3,2} & [A]_{3,3} \\
                  \end{bmatrix}}
        + [A]_{1,1}
        \mathrm{D} \left(
        \det {\begin{bmatrix}
                      [A]_{2,2} & [A]_{2,3} \\
                      [A]_{3,2} & [A]_{3,3} \\
                  \end{bmatrix}}
        \right)
        \\
             &
        - \mathrm{D} ([A]_{2,1})
        \det {\begin{bmatrix}
                      [A]_{1,2} & [A]_{1,3} \\
                      [A]_{3,2} & [A]_{3,3} \\
                  \end{bmatrix}}
        - [A]_{2,1}
        \mathrm{D} \left(
        \det {\begin{bmatrix}
                      [A]_{1,2} & [A]_{1,3} \\
                      [A]_{3,2} & [A]_{3,3} \\
                  \end{bmatrix}}
        \right)
        \\
             &
        + \mathrm{D} ([A]_{3,1})
        \det {\begin{bmatrix}
                      [A]_{1,2} & [A]_{1,3} \\
                      [A]_{2,2} & [A]_{2,3} \\
                  \end{bmatrix}}
        + [A]_{3,1}
        \mathrm{D} \left(
        \det {\begin{bmatrix}
                      [A]_{1,2} & [A]_{1,3} \\
                      [A]_{2,2} & [A]_{2,3} \\
                  \end{bmatrix}}
        \right)
        \\
        = {} &
        \hphantom{{} + {}}
        \mathrm{D} ([A]_{1,1})
        \det {\begin{bmatrix}
                      [A]_{2,2} & [A]_{2,3} \\
                      [A]_{3,2} & [A]_{3,3} \\
                  \end{bmatrix}}
        + [A]_{1,1}
        \det {\begin{bmatrix}
                      \mathrm{D} ([A]_{2,2}) & [A]_{2,3} \\
                      \mathrm{D} ([A]_{3,2}) & [A]_{3,3} \\
                  \end{bmatrix}}
        \\
             &
        + [A]_{1,1}
        \det {\begin{bmatrix}
                      [A]_{2,2} & \mathrm{D} ([A]_{2,3}) \\
                      [A]_{3,2} & \mathrm{D} ([A]_{3,3}) \\
                  \end{bmatrix}}
        \\
             &
        - \mathrm{D} ([A]_{2,1})
        \det {\begin{bmatrix}
                      [A]_{1,2} & [A]_{1,3} \\
                      [A]_{3,2} & [A]_{3,3} \\
                  \end{bmatrix}}
        - [A]_{2,1}
        \det {\begin{bmatrix}
                      \mathrm{D} ([A]_{1,2}) & [A]_{1,3} \\
                      \mathrm{D} ([A]_{3,2}) & [A]_{3,3} \\
                  \end{bmatrix}}
        \\
             &
        - [A]_{2,1}
        \det {\begin{bmatrix}
                      [A]_{1,2} & \mathrm{D} ([A]_{1,3}) \\
                      [A]_{3,2} & \mathrm{D} ([A]_{3,3}) \\
                  \end{bmatrix}}
        \\
             &
        + \mathrm{D} ([A]_{3,1})
        \det {\begin{bmatrix}
                      [A]_{1,2} & [A]_{1,3} \\
                      [A]_{2,2} & [A]_{2,3} \\
                  \end{bmatrix}}
        + [A]_{3,1}
        \det {\begin{bmatrix}
                      \mathrm{D} ([A]_{1,2}) & [A]_{1,3} \\
                      \mathrm{D} ([A]_{2,2}) & [A]_{2,3} \\
                  \end{bmatrix}}
        \\
             &
        + [A]_{3,1}
        \det {\begin{bmatrix}
                      [A]_{1,2} & \mathrm{D} ([A]_{1,3}) \\
                      [A]_{2,2} & \mathrm{D} ([A]_{2,3}) \\
                  \end{bmatrix}}
        \\
        = {} &
        \hphantom{{} + {}}
        \det {\begin{bmatrix}
                      \mathrm{D} ([A]_{1,1}) & [A]_{1,2} & [A]_{1,3} \\
                      \mathrm{D} ([A]_{2,1}) & [A]_{2,2} & [A]_{2,3} \\
                      \mathrm{D} ([A]_{3,1}) & [A]_{3,2} & [A]_{3,3} \\
                  \end{bmatrix}}
        + \det {\begin{bmatrix}
                        [A]_{1,1} & \mathrm{D} ([A]_{1,2}) & [A]_{1,3} \\
                        [A]_{2,1} & \mathrm{D} ([A]_{2,2}) & [A]_{2,3} \\
                        [A]_{3,1} & \mathrm{D} ([A]_{3,2}) & [A]_{3,3} \\
                    \end{bmatrix}}
        \\
             &
        + \det {\begin{bmatrix}
                        [A]_{1,1} & [A]_{1,2} & \mathrm{D} ([A]_{1,3}) \\
                        [A]_{2,1} & [A]_{2,2} & \mathrm{D} ([A]_{2,3}) \\
                        [A]_{3,1} & [A]_{3,2} & \mathrm{D} ([A]_{3,3}) \\
                    \end{bmatrix}}
        \,.
    \end{align*}

    我们想, 我们或许知道 \(\mathrm{D} (\det {(A)})\) 了.
    为方便, 我们作这样的记号:
    若 \(B\) 是 \(m \times n\)~阵,
    且 \(\ell\) 是正整数,
    则 \(\mathrm{D}_\ell (B)\) 也是 \(m \times n\)~阵,
    其中,
    \begin{align*}
        [\mathrm{D}_\ell (B)]_{i,j} =
        \begin{cases}
            \mathrm{D} ([B]_{i,j}), & j = \ell;    \\
            [B]_{i,j},              & \text{别的情形}.
        \end{cases}
    \end{align*}

    作命题 \(P(n)\):
    对任何 \(n\)~级阵 \(A\),
    \begin{align*}
        \mathrm{D} (\det {(A)})
        = \sum_{\ell = 1}^{n} {\det {(\mathrm{D}_\ell (A))}}.
    \end{align*}
    我们的目标是,
    对任何正整数 \(n\), \(P(n)\) 是对的.

    \(P(1)\) 与 \(P(2)\) 是对的.

    假定 \(P(n-1)\) 是对的.
    我们由此证明, \(P(n)\) 也是对的.

    注意,
    \begin{align*}
             &
        \mathrm{D} (\det {(A)})
        \\
        = {} &
        \mathrm{D} \left(
        \sum_{i = 1}^{n} {(-1)^{i+1} [A]_{i,1} \det {(A(i|1))}}
        \right)
        \\
        = {} &
        \sum_{i = 1}^{n} {(-1)^{i+1} \mathrm{D} ([A]_{i,1} \det {(A(i|1))})}
        \\
        = {} &
        \sum_{i = 1}^{n} {(-1)^{i+1}
        (\mathrm{D} ([A]_{i,1}) \det {(A(i|1))}
        + [A]_{i,1} \mathrm{D} (\det {(A(i|1))}))}
        \\
        = {} &
        \sum_{i = 1}^{n} {(-1)^{i+1}
                \mathrm{D} ([A]_{i,1}) \det {(A(i|1))}}
        +
        \sum_{i = 1}^{n} {(-1)^{i+1}
                    [A]_{i,1} \mathrm{D} (\det {(A(i|1))})}.
    \end{align*}
    注意, 既然 \(A\) 与 \(\mathrm{D}_1 (A)\)
    的列~\(j\) 相同 (若 \(j \neq 1\)),
    则 \(A(i|1) = (\mathrm{D}_1 (A))(i|1)\).
    则
    \begin{align*}
             &
        \sum_{i = 1}^{n} {(-1)^{i+1}
                \mathrm{D} ([A]_{i,1}) \det {(A(i|1))}}
        \\
        = {} &
        \sum_{i = 1}^{n} {(-1)^{i+1}
                \mathrm{D} ([A]_{i,1}) \det {((\mathrm{D}_1 (A))(i|1))}}
        \\
        = {} &
        \det {(\mathrm{D}_1 (A))}.
    \end{align*}

    由假定,
    \begin{align*}
        \mathrm{D} (\det {(A(i|1))}) =
        \sum_{v = 1}^{n-1} {\det {(\mathrm{D}_v (A(i|1)))}}.
    \end{align*}
    注意, \(\mathrm{D}_v (A(i|1))
    = (\mathrm{D}_{v+1} (A)) (i|1)\).
    则
    \begin{align*}
        \mathrm{D} (\det {(A(i|1))})
        =
        \sum_{v = 1}^{n-1} {\det {((\mathrm{D}_{v+1} (A)) (i|1))}}
        =
        \sum_{\ell = 2}^{n} {\det {((\mathrm{D}_\ell (A)) (i|1))}}.
    \end{align*}
    注意, 当 \(\ell \neq 1\) 时,
    \([\mathrm{D}_\ell (A)]_{i,1} = [A]_{i,1}\).
    则
    \begin{align*}
             &
        \sum_{i = 1}^{n} {(-1)^{i+1}
                    [A]_{i,1} \mathrm{D} (\det {(A(i|1))})}
        \\
        = {} &
        \sum_{i = 1}^{n} {(-1)^{i+1}
                    [\mathrm{D}_\ell (A)]_{i,1}
                \sum_{\ell = 2}^{n} {\det {((\mathrm{D}_\ell (A)) (i|1))}}
            }
        \\
        = {} &
        \sum_{i = 1}^{n} {
                \sum_{\ell = 2}^{n} {
                        (-1)^{i+1} [\mathrm{D}_\ell (A)]_{i,1}
                        \det {((\mathrm{D}_\ell (A)) (i|1))}
                    }
            }
        \\
        = {} &
        \sum_{\ell = 2}^{n} {
                \sum_{i = 1}^{n} {
                        (-1)^{i+1} [\mathrm{D}_\ell (A)]_{i,1}
                        \det {((\mathrm{D}_\ell (A)) (i|1))}
                    }
            }
        \\
        = {} &
        \sum_{\ell = 2}^{n} {\det {(\mathrm{D}_\ell (A))}}.
    \end{align*}
    综上,
    \begin{align*}
        \mathrm{D} (\det {(A)})
        = \det {(\mathrm{D}_1 (A))}
        + \sum_{\ell = 2}^{n} {\det {(\mathrm{D}_\ell (A))}}
        = \sum_{\ell = 1}^{n} {\det {(\mathrm{D}_\ell (A))}}.
    \end{align*}
    所以, \(P(n)\) 是对的.
    由数学归纳法, 待证命题成立.

    注意, 既然 \(A\) 与 \(\mathrm{D}_\ell (A)\)
    的列~\(j\) 相同 (若 \(j \neq \ell\)),
    则 \(A(i|\ell) = (\mathrm{D}_\ell (A))(i|\ell)\).
    并且,
    \([\mathrm{D}_\ell (A)]_{i,\ell} = \mathrm{D} ([A]_{i,\ell})\).
    则
    \begin{align*}
        \mathrm{D} (\det {(A)})
        = {} &
        \sum_{\ell = 1}^{n} {\det {(\mathrm{D}_\ell (A))}}
        \\
        = {} &
        \sum_{\ell = 1}^{n} {
                \sum_{i = 1}^{n} {
                        (-1)^{i+\ell} [\mathrm{D}_\ell (A)]_{i,\ell}
                        \det {((\mathrm{D}_\ell (A))(i|\ell))}
                    }
            }
        \\
        = {} &
        \sum_{\ell = 1}^{n} {
                \sum_{i = 1}^{n} {
                        (-1)^{i+\ell} \mathrm{D} ([A]_{i,\ell})
                        \det {(A(i|\ell))}
                    }
            }
        \\
        = {} &
        \sum_{\ell = 1}^{n} {
                \sum_{i = 1}^{n} {
                        (-1)^{i+\ell}
                        \det {(A(i|\ell))}\,
                        \mathrm{D} ([A]_{i,\ell})
                    }
            }
        \\
        = {} &
        \sum_{\ell = 1}^{n} {
                \sum_{i = 1}^{n} {
                        [\operatorname{adj} {(A)}]_{\ell,i}\,
                        \mathrm{D} ([A]_{i,\ell})
                    }
            }.
    \end{align*}
    为方便, 我们作这样的记号:
    若 \(B\) 是 \(m \times n\)~阵,
    则 \(\mathrm{D}_{\mathrm{c}} (B)\) 也是 \(m \times n\)~阵,
    其中,
    \([\mathrm{D}_{\mathrm{c}} (B)]_{i,j} = \mathrm{D} ([B]_{i,j})\).
    则
    \begin{align*}
        \mathrm{D} (\det {(A)})
        = \sum_{\ell = 1}^{n}
            {[\operatorname{adj} {(A)}\,
                        \mathrm{D}_{\mathrm{c}} (A)]_{\ell,\ell}}.
    \end{align*}

    最后, 我们再引入方阵的 trace:
    若 \(A\) 是 \(n\)~级阵, 则 \(A\) 的
    \emph{trace} \pinjino{chuisi}
    \begin{align*}
        \operatorname{tr} {(A)} = \sum_{k = 1}^{n} {[A]_{k,k}}.
    \end{align*}
    则
    \begin{align*}
        \mathrm{D} (\det {(A)})
        = \operatorname{tr} {(\operatorname{adj} {(A)}\,
            \mathrm{D}_{\mathrm{c}} (A))}.
    \end{align*}
\end{example}

\begin{example}
    设 \(A\) 是 \(n\)~级反称阵.
    我们求 \(\mathrm{D} (\operatorname{pf} {(A)})\).

    设 \(n = 1\).
    则 \(\operatorname{pf} {(A)} = 0\).
    则 \(\mathrm{D} (\operatorname{pf} {(A)}) = 0\).

    设 \(n = 2\).
    则 \(\operatorname{pf} {(A)} = [A]_{1,2}\).
    则 \(\mathrm{D} (\operatorname{pf} {(A)})
    = \mathrm{D} ([A]_{1,2})\).

    设 \(n = 3\).
    则 \(\operatorname{pf} {(A)} = 0\).
    则 \(\mathrm{D} (\operatorname{pf} {(A)}) = 0\).

    设 \(n = 4\).
    则 \(\operatorname{pf} {(A)}
    = [A]_{1,2} [A]_{3,4} - [A]_{1,3} [A]_{2,4} + [A]_{1,4} [A]_{2,3}\).
    则
    \begin{align*}
             &
        \mathrm{D} (\operatorname{pf} {(A)})
        \\
        = {} &
        \hphantom{{} + {}}
        \mathrm{D} ([A]_{1,2})\, [A]_{3,4}
        + [A]_{1,2}\, \mathrm{D} ([A]_{3,4})
        - \mathrm{D} ([A]_{1,3})\, [A]_{2,4}
        - [A]_{1,3}\, \mathrm{D} ([A]_{2,4})
        \\
             &
        + \mathrm{D} ([A]_{1,4})\, [A]_{2,3}
        + [A]_{1,4}\, \mathrm{D} ([A]_{2,3})
        \\
        = {} &
        \hphantom{{} + {}}
        \mathrm{D} ([A]_{1,2})\, [A]_{3,4}
        - \mathrm{D} ([A]_{1,3})\, [A]_{2,4}
        + \mathrm{D} ([A]_{1,4})\, [A]_{2,3}
        \\
             &
        + \mathrm{D} ([A]_{2,3})\, [A]_{1,4}
        - \mathrm{D} ([A]_{2,4})\, [A]_{1,3}
        + \mathrm{D} ([A]_{3,4})\, [A]_{1,2}.
    \end{align*}

    作命题 \(P(n)\):
    对任何 \(n\)~级反称阵 \(A\),
    \begin{align*}
        \mathrm{D} (\operatorname{pf} {(A)})
        = \sum_{1 \leq i < j \leq n}
        {(-1)^{i+j-1} \mathrm{D} ([A]_{i,j})
        \operatorname{pf} {(A({i,j}|{i,j}))}}.
    \end{align*}
    再作命题 \(Q(n)\):
    \(P(n-1)\) 与 \(P(n)\) 是对的.
    我们用数学归纳法证明,
    对任何大于 \(3\) 的正整数 \(n\), \(Q(n)\) 是对的.
    则对不低于 \(3\) 的正整数 \(n\), \(P(n)\) 是对的.

    \(P(3)\) 与 \(P(4)\) 显然地是对的.
    则 \(Q(4)\) 是对的.

    设 \(Q(n-1)\) 是对的.
    则 \(P(n-2)\) 与 \(P(n-1)\) 是对的.
    我们要证, \(Q(n)\) 是对的.
    则我们要证, \(P(n-1)\) 与 \(P(n)\) 是对的.
    于是, 若我们以 \(P(n-2)\) 与 \(P(n-1)\) 证,
    \(P(n)\) 是对的,
    则 \(Q(n)\) 是对的.

    设 \(n > 4\).
    则
    \begin{align*}
             &
        \mathrm{D} (\operatorname{pf} {(A)})
        \\
        = {} &
        \mathrm{D} \left(
        \sum_{2 \leq \ell \leq n}
        {(-1)^{\ell} [A]_{1,\ell}
        \operatorname{pf} {(A({1,\ell}|{1,\ell}))}}
        \right)
        \\
        = {} &
        \sum_{2 \leq \ell \leq n}
        {(-1)^{\ell}
        \mathrm{D} (
        [A]_{1,\ell}
        \operatorname{pf} {(A({1,\ell}|{1,\ell}))})
        }
        \\
        = {} &
        \sum_{2 \leq \ell \leq n}
        {(-1)^{\ell}
        (
        \mathrm{D} ([A]_{1,\ell})
        \operatorname{pf} {(A({1,\ell}|{1,\ell}))}
        +
        [A]_{1,\ell}
        \mathrm{D} (\operatorname{pf} {(A({1,\ell}|{1,\ell}))})
        )
        }
        \\
        = {} &
        \hphantom{{} + {}}
        \sum_{2 \leq \ell \leq n}
        {(-1)^{\ell}
        \mathrm{D} ([A]_{1,\ell})
        \operatorname{pf} {(A({1,\ell}|{1,\ell}))}
        }
        \\
             & +
        \sum_{2 \leq \ell \leq n}
        {(-1)^{\ell} [A]_{1,\ell}
        \mathrm{D} (\operatorname{pf} {(A({1,\ell}|{1,\ell}))})
        }.
    \end{align*}
    注意, \([A]_{i,j}\) 在 \(A({1,\ell}|{1,\ell})\)
    的位置是行~\(i - \rho(i, 1) - \rho(i, \ell)\),
    列~\(j - \rho(j, 1) - \rho(j, \ell)\).
    (回想,
    若 \(i \geq j\), 则 \(\rho(i, j) = 1\),
    且若 \(i < j\), 则 \(\rho(i, j) = 0\).)
    由假定,
    \begin{align*}
             &
        \mathrm{D} (\operatorname{pf} {(A({1,\ell}|{1,\ell}))})
        \\
        = {} &
        \sum_{\substack{2 \leq i < j \leq n \\ i, j \neq \ell}}
        {(-1)^{i-1-\rho(i, \ell)+j-1-\rho(j, \ell)-1}
        \mathrm{D} ([A]_{i,j})
        \operatorname{pf} {(A({1,\ell,i,j}|{1,\ell,i,j}))}}
        \\
        = {} &
        \sum_{\substack{2 \leq i < j \leq n \\ i, j \neq \ell}}
        {(-1)^{i-1+(\rho(\ell, i)-1)+j-1+(\rho(\ell, j)-1)-1}
        \mathrm{D} ([A]_{i,j})
        \operatorname{pf} {(A({1,\ell,i,j}|{1,\ell,i,j}))}}
        \\
        = {} &
        \sum_{\substack{2 \leq i < j \leq n \\ i, j \neq \ell}}
        {(-1)^{(i+j-1)-1-\rho(\ell, i)+1-1-\rho(\ell, j)+1}
        \mathrm{D} ([A]_{i,j})
        \operatorname{pf} {(A({1,\ell,i,j}|{1,\ell,i,j}))}}
        \\
        = {} &
        \sum_{\substack{2 \leq i < j \leq n \\ i, j \neq \ell}}
        {(-1)^{(i+j-1)-\rho(\ell, i)-\rho(\ell, j)}
        \mathrm{D} ([A]_{i,j})
        \operatorname{pf} {(A({1,\ell,i,j}|{1,\ell,i,j}))}}.
    \end{align*}
    为方便, 我们简单地写 \(A({1,\ell,i,j}|{1,\ell,i,j})\)
    为 \(A_{1,\ell,i,j}\).
    则
    \begin{align*}
             &
        \sum_{2 \leq \ell \leq n}
        (-1)^{\ell} [A]_{1,\ell}
        \mathrm{D} (\operatorname{pf} {(A({1,\ell}|{1,\ell}))})
        \\
        = {} &
        \sum_{2 \leq \ell \leq n}
        (-1)^{\ell} [A]_{1,\ell}
        \sum_{\substack{2 \leq i < j \leq n \\ i, j \neq \ell}}
        (-1)^{(i+j-1)-\rho(\ell, i)-\rho(\ell, j)}
        \mathrm{D} ([A]_{i,j})
        \operatorname{pf} {(A_{1,\ell,i,j})}
        \\
        = {} &
        \sum_{2 \leq \ell \leq n}
        \sum_{\substack{2 \leq i < j \leq n \\ i, j \neq \ell}}
        (-1)^{\ell} [A]_{1,\ell}
        (-1)^{(i+j-1)-\rho(\ell, i)-\rho(\ell, j)}
        \mathrm{D} ([A]_{i,j})
        \operatorname{pf} {(A_{1,\ell,i,j})}
        \\
        = {} &
        \sum_{2 \leq \ell \leq n}
        \sum_{\substack{2 \leq i < j \leq n \\ i, j \neq \ell}}
        (-1)^{i+j-1} \mathrm{D} ([A]_{i,j})
        (-1)^{\ell-\rho(\ell, i)-\rho(\ell, j)} [A]_{1,\ell}
        \operatorname{pf} {(A_{1,\ell,i,j})}
        \\
        = {} &
        \sum_{2 \leq i < j \leq n}
        \sum_{\substack{2 \leq \ell \leq n  \\ \ell \neq i, j}}
        (-1)^{i+j-1} \mathrm{D} ([A]_{i,j})
        (-1)^{\ell-\rho(\ell, i)-\rho(\ell, j)} [A]_{1,\ell}
        \operatorname{pf} {(A_{1,\ell,i,j})}
        \\
        = {} &
        \sum_{2 \leq i < j \leq n}
        (-1)^{i+j-1} \mathrm{D} ([A]_{i,j})
        \sum_{\substack{2 \leq \ell \leq n  \\ \ell \neq i, j}}
        (-1)^{\ell-\rho(\ell, i)-\rho(\ell, j)} [A]_{1,\ell}
        \operatorname{pf} {(A_{1,\ell,i,j})}
        \\
        = {} &
        \sum_{\substack{1 \leq i < j \leq n \\ i \geq 2}}
        {
        (-1)^{i+j-1} \mathrm{D} ([A]_{i,j})
        \operatorname{pf} {(A({i,j}|{i,j}))}
        }.
    \end{align*}
    注意,
    \begin{align*}
             &
        \sum_{2 \leq \ell \leq n}
        {(-1)^{\ell}
        \mathrm{D} ([A]_{1,\ell})
        \operatorname{pf} {(A({1,\ell}|{1,\ell}))}
        }
        \\
        = {} &
        \sum_{\substack{1 \leq i < j \leq n \\ i = 1}}
        {
        (-1)^{i+j-1} \mathrm{D} ([A]_{i,j})
        \operatorname{pf} {(A({i,j}|{i,j}))}
        }.
    \end{align*}
    综上,
    \begin{align*}
        \mathrm{D} (\operatorname{pf} {(A)})
        = {} &
        \hphantom{{} + {}}
        \sum_{\substack{1 \leq i < j \leq n \\ i = 1}}
        {
        (-1)^{i+j-1} \mathrm{D} ([A]_{i,j})
        \operatorname{pf} {(A({i,j}|{i,j}))}
        }
        \\
             &
        +
        \sum_{\substack{1 \leq i < j \leq n \\ i \geq 2}}
        {
        (-1)^{i+j-1} \mathrm{D} ([A]_{i,j})
        \operatorname{pf} {(A({i,j}|{i,j}))}
        }
        \\
        = {} &
        \sum_{1 \leq i < j \leq n}
        {(-1)^{i+j-1} \mathrm{D} ([A]_{i,j})
        \operatorname{pf} {(A({i,j}|{i,j}))}}.
    \end{align*}
    所以, \(P(n)\) 是对的.
    则 \(Q(n)\) 是对的.
    由数学归纳法, 待证命题成立.
\end{example}

\begin{example}
    回想, 若 \(A\) 是 \(n\)~级阵, 则 \(A\) 的 trace
    \begin{align*}
        \operatorname{tr} {(A)} = \sum_{k = 1}^{n} {[A]_{k,k}}.
    \end{align*}
    我们说, trace 也是方阵的一个属性,
    就像行列式是方阵的一个属性那样.

    设 \(A\), \(B\) 是 \(n\)~级阵.
    设 \(s\) 是数.
    设 \(C\) 是 \(n \times m\)~阵,
    且 \(D\) 是 \(m \times n\)~阵.
    则:

    (1)
    \(\operatorname{tr} {(A + B)}
    = \operatorname{tr} {(A)} + \operatorname{tr} {(B)}\).
    注意,
    \begin{align*}
        \operatorname{tr} {(A + B)}
        = {} &
        \sum_{k = 1}^{n} {[A + B]_{k,k}}
        \\
        = {} &
        \sum_{k = 1}^{n} {([A]_{k,k} + [B]_{k,k})}
        \\
        = {} &
        \sum_{k = 1}^{n} {[A]_{k,k}}
        + \sum_{k = 1}^{n} {[B]_{k,k}}
        \\
        = {} &
        \operatorname{tr} {(A)}
        + \operatorname{tr} {(B)}.
    \end{align*}

    (2)
    \(\operatorname{tr} {(s A)} = s \operatorname{tr} {(A)}\).
    注意,
    \begin{align*}
        \operatorname{tr} {(s A)}
        = \sum_{k = 1}^{n} {[s A]_{k,k}}
        = \sum_{k = 1}^{n} {s\, [A]_{k,k}}
        = s \sum_{k = 1}^{n} {[A]_{k,k}}
        = s \operatorname{tr} {(A)}.
    \end{align*}

    (3)
    \(\operatorname{tr} {(A^{\mathrm{T}})} = \operatorname{tr} {(A)}\).
    注意,
    \begin{align*}
        \operatorname{tr} {(A^{\mathrm{T}})}
        = \sum_{k = 1}^{n} {[A^{\mathrm{T}}]_{k,k}}
        = \sum_{k = 1}^{n} {[A]_{k,k}}
        = \operatorname{tr} {(A)}.
    \end{align*}

    (4)
    \(\operatorname{tr} {(C D)} = \operatorname{tr} {(D C)}\).
    注意,
    \begin{align*}
        \operatorname{tr} {(C D)}
        = {} &
        \sum_{k = 1}^{n} {[C D]_{k,k}}
        \\
        = {} &
        \sum_{k = 1}^{n} {
                \sum_{\ell = 1}^{m} {
                        [C]_{k,\ell} [D]_{\ell,k}
                    }
            }
        \\
        = {} &
        \sum_{\ell = 1}^{m} {
                \sum_{k = 1}^{n} {
                        [C]_{k,\ell} [D]_{\ell,k}
                    }
            }
        \\
        = {} &
        \sum_{\ell = 1}^{m} {
                \sum_{k = 1}^{n} {
                        [D]_{\ell,k} [C]_{k,\ell}
                    }
            }
        \\
        = {} &
        \sum_{\ell = 1}^{m} {[D C]_{\ell,\ell}}
        \\
        = {} &
        \operatorname{tr} {(D C)}.
    \end{align*}
    (其实, 此事是例~\ref{emp:sum-of-principal-minors-of-AB-and-that-of-BA}
    的一个特别的情形.)

    值得提, 当 \(n \geq 2\) 时,
    \(\operatorname{tr} {(A B)}
    = \operatorname{tr} {(A)} \operatorname{tr} {(B)}\)
    一般不是对的.
    为此, 设 \(A\), \(B\) 是这样的 \(n\)~级阵:
    \begin{align*}
        [A]_{i,j} =
        \begin{cases}
            1, & i = j = 1;   \\
            0, & \text{别的情形},
        \end{cases}
        \quad
        [B]_{i,j} =
        \begin{cases}
            1, & i = j = 2;   \\
            0, & \text{别的情形}.
        \end{cases}
    \end{align*}
    可以验证, \(A B = 0\).
    则 \(\operatorname{tr} {(A B)} = 0\).
    但是, \(\operatorname{tr} {(A)} = 1 = \operatorname{tr} {(B)}\).
\end{example}

\begin{example}
    设 \(A\) 是 \(n\)~级阵.
    则存在 \(n\)~级阵 \(B\),
    使 \(AB = \det {(A)}\, I_n = BA\).

    我们知道, \(A \operatorname{adj} {(A)}
    = \det {(A)}\, I_n
    = \operatorname{adj} {(A)}\, A\).
    若 \(\operatorname{adj} {(A)} \neq 0\),
    我们取 \(B\) 为 \(\operatorname{adj} {(A)}\).

    以下, 我们设 \(\operatorname{adj} {(A)} = 0\).
    注意, 此时, \(n > 1\)
    (注意, 若 \(n = 1\), 则 \(A\) 的古伴总为 \(I_1\)),
    且 \(\det {(A)} = 0\).

    若 \(A = 0\), 我们可取 \(B\) 为任何非零的 \(n\)~级阵
    (如 \(I_n\)).

    以下, 我们设 \(\operatorname{adj} {(A)} = 0\),
    且 \(A \neq 0\).
    注意, 此时, \(n > 2\)
    (注意, 若 \(n = 2\), 则 \(A\) 的古伴不为 \(0\),
    除非 \(A = 0\)).

    我们说, 存在小于 \(n-1\) 的正整数 \(r\),
    使 \(A\) 有行列式非零的 \(r\)~级子阵,
    且 \(A\) 没有行列式非零的 \(r+1\)~级子阵.
    设
    \(\displaystyle
    C = A\binom{i_1,i_2,\dots,i_{r+1}}{j_1,j_2,\dots,j_{r+1}}\)
    (其中, \(i_1 < i_2 < \dots < i_{r+1}\),
    且 \(j_1 < j_2 < \dots < j_{r+1}\);
    则 \([C]_{s,t} = [A]_{i_s,j_t}\))
    是 \(A\) 的一个 \(r+1\)~级子阵,
    且 \(C\) 有行列式非零的 \(r\)~级子阵
    (此 \(r\)~级子阵当然是 \(A\) 的行列式非零的 \(r\)~级子阵).
    则 \(\operatorname{adj} {(C)} \neq 0\),
    且 \(C \operatorname{adj} {(C)} = 0
    = \operatorname{adj} {(C)}\, C\).

    作 \(n\)~级阵 \(B\), 使
    \begin{align*}
        [B]_{i,j} =
        \begin{cases}
            [\operatorname{adj} {(C)}]_{t,s},
             & \text{若 \(i\) 等于某 \(j_t\), 且 \(j\) 等于某 \(i_s\)};
            \\
            0,
             & \text{别的情形}.
        \end{cases}
    \end{align*}
    则 \(B \neq 0\)
    (因为 \(\operatorname{adj} {(C)}\) 的元都是 \(B\) 的元,
    且 \(\operatorname{adj} {(C)} \neq 0\)).
    我们说明, \(A B = 0 = \det {(A)}\, I_n\),
    且 \(B A = 0 = \det {(A)}\, I_n\).

    注意,
    \begin{align*}
        [A B]_{i,j}
        = {} &
        \sum_{1 \leq \ell \leq n} {[A]_{i,\ell} [B]_{\ell,j}}
        \\
        = {} &
        \sum_{\substack{1 \leq \ell \leq n   \\ \ell\,\text{等于某}\,j_t}}
        {[A]_{i,\ell} [B]_{\ell,j}}
        + \sum_{\substack{1 \leq \ell \leq n \\ \ell\,\text{不等于任何}\,j_t}}
        {[A]_{i,\ell} [B]_{\ell,j}}
        \\
        = {} &
        \sum_{1 \leq t \leq r+1}
        {[A]_{i,j_t} [B]_{j_t,j}}
        + \sum_{\substack{1 \leq \ell \leq n \\ \ell\,\text{不等于任何}\,j_t}}
        {[A]_{i,\ell} 0}
        \\
        = {} &
        \sum_{1 \leq t \leq r+1}
        {[A]_{i,j_t} [B]_{j_t,j}}.
    \end{align*}
    若 \(j\) 不等于任何 \(i_s\), 则 \([B]_{j_t,j} = 0\).
    则 \([A B]_{i,j} = 0\).
    若 \(j\) 等于某 \(i_s\), 则
    \begin{align*}
        [A B]_{i,i_s}
        = {} &
        \sum_{1 \leq t \leq r+1} {[A]_{i,j_t} [B]_{j_t,i_s}}
        \\
        = {} &
        \sum_{1 \leq t \leq r+1}
        {[A]_{i,j_t} [\operatorname{adj} {(C)}]_{t,s}}.
    \end{align*}
    若 \(i\) 等于某 \(i_k\), 则
    \begin{align*}
        [A B]_{i_k,i_s}
        = {} &
        \sum_{1 \leq t \leq r+1}
        {[A]_{i_k,j_t} [\operatorname{adj} {(C)}]_{t,s}}
        \\
        = {} &
        \sum_{1 \leq t \leq r+1}
        {[C]_{k,t} [\operatorname{adj} {(C)}]_{t,s}}
        \\
        = {} &
        [C \operatorname{adj} {(C)}]_{k,s}
        \\
        = {} &
        [0]_{k,s}
        \\
        = {} &
        0.
    \end{align*}
    若 \(i\) 不等于任何 \(i_k\), 则
    \begin{align*}
        [A B]_{i,i_s}
        = {} &
        \sum_{1 \leq t \leq r+1}
        {[A]_{i,j_t} [\operatorname{adj} {(C)}]_{t,s}}
        \\
        = {} &
        \sum_{1 \leq t \leq r+1}
        {[A]_{i,j_t}\, (-1)^{s+t} \det {(C(s|t))}}.
    \end{align*}
    作 \(r+1\)~级阵 \(D\), 使
    \begin{align*}
        [D]_{u,v} =
        \begin{cases}
            [A]_{i,j_v}, & u = s;       \\
            [C]_{u,v},   & \text{别的情形}.
        \end{cases}
    \end{align*}
    则 \(C\) 的行 \(u\) 与 \(D\) 的行 \(u\) 相等
    (若 \(u \neq s\)).
    则 \(C(s|t) = D(s|t)\),
    且 \([A]_{i,j_t} = [D]_{s,t}\).
    则 \([A B]_{i,i_s} = \det {(D)}\).

    设 \(i\) 在
    \(i_1\), \(\dots\), \(i_{s-1}\), \(i\), \(i_{s+1}\), \(\dots\), \(i_{r+1}\)
    中是第 \(z\) 小的数.
    若 \(z = s\), 则 \(D\) 就是 \(A\) 的 \(r+1\)~级子阵
    \(\displaystyle
    F = A\binom{i_1,\dots,i_{s-1},i,i_{s+1},\dots,i_{r+1}}
    {j_1,j_2,\dots,j_{r+1}}\).
    注意, \(\det {(F)} = 0\).
    则 \(\det {(D)} = 0\).
    若 \(z > s\),
    则我们分别地使 \(D\) 的行~\(s\)
    与在行~\(s\) 的下面的 \(z - s\) 行,
    行~\(i_{s+1}\), \(\dots\), 行~\(i_z\),
    交换位置, 变 \(D\) 为 \(F\).
    则 \(\det {(D)} = (-1)^{z - s} \det {(F)} = 0\).
    若 \(z < s\),
    则我们分别地使 \(D\) 的行~\(s\)
    与在行~\(s\) 的上面的 \(s - z\) 行,
    行~\(i_{s-1}\), \(\dots\), 行~\(i_z\),
    交换位置, 变 \(D\) 为 \(F\).
    则 \(\det {(D)} = (-1)^{s - z} \det {(F)} = 0\).
    则 \([A B]_{i,i_s} = \det {(D)} = 0\).

    综上, 既然 \(A B\) 的每个元都是 \(0\), 则 \(A B = 0\).

    我们可几乎类似地证, \(B A = 0\).
    当然, 我们也可这样作.

    \(B A = 0\) 相当于
    \(A^{\mathrm{T}} B^{\mathrm{T}} = 0^{\mathrm{T}} = 0\).
    注意, \(A^{\mathrm{T}}\) 也有行列式非零的 \(r\)~级子阵,
    且 \(A^{\mathrm{T}}\) 也没有行列式非零的 \(r+1\)~级子阵.
    注意,
    \(\displaystyle
    C^{\mathrm{T}}
    = A^{\mathrm{T}}
    \binom{j_1,j_2,\dots,j_{r+1}}{i_1,i_2,\dots,i_{r+1}}\)
    是 \(A^{\mathrm{T}}\) 的一个 \(r+1\)~级子阵,
    且 \(C^{\mathrm{T}}\) 有行列式非零的 \(r\)~级子阵
    (此 \(r\)~级子阵当然是 \(A^{\mathrm{T}}\) 的行列式非零的 \(r\)~级子阵).
    则 \(\operatorname{adj} {(C^{\mathrm{T}})} \neq 0\),
    且 \(C^{\mathrm{T}} \operatorname{adj} {(C^{\mathrm{T}})} = 0
    = \operatorname{adj} {(C^{\mathrm{T}})}\, C^{\mathrm{T}}\).

    我们类似地作 \(n\)~级阵 \(G\), 使
    \begin{align*}
        [G]_{i,j} =
        \begin{cases}
            [\operatorname{adj} {(C^{\mathrm{T}})}]_{s,t},
             & \text{若 \(i\) 等于某 \(i_s\), 且 \(j\) 等于某 \(j_t\)};
            \\
            0,
             & \text{别的情形}.
        \end{cases}
    \end{align*}
    则 \(G = B^{\mathrm{T}}\).
    (注意, \([\operatorname{adj} {(C^{\mathrm{T}})}]_{s,t}
        = [\operatorname{adj} {(C)}]_{t,s}\),
    故 \([G]_{i_s,j_t} = [B]_{j_t,i_s}\).
    在别的情形, \([G]_{i,j} = 0 = [B]_{j,i}\).)
    并且, 当然, \(G \neq 0\).
    那么, 由前面的讨论, \(A^{\mathrm{T}} B^{\mathrm{T}} = 0\).
\end{example}

\end{document}

This material contains many miscellaneous examples
about determinants
(which seem to be hard for beginners):
\begin{itemize}
    \item a formula for det (D + A) in which D is diagonal
    \item a formula for det (xI + A)
    \item a formula that relates det (xI + AB) and det (xI + BA)
    \item powers of a square matrix
    \item a formula for adj (-A) in terms of a polynomial in A
    \item Cayley--Hamilton
    \item more about skew-symmetric matrices
    \item the derivative of the determinant of a square matrix
    \item the trace of a square matrix
\end{itemize}
This material consists of one very long section.

This example was rewritten, and it was moved to another place:

\begin{example}
    设定义在全体 \(n\)~级阵上的函数 \(f\) 适合:

    (1)
    \(f(A B) = f(A) f(B)\),
    对任何 \(n\)~级阵 \(A\), \(B\);

    (2)
    若 \(n\)~级阵 \(D\) 适合
    \([D]_{i,j} = 0\) (\(i \neq j\)),
    则 \(f(D) = \det {(D)}\).
    % (2)
    % 若 \(n\)~级阵 \(D\) 适合 \([D]_{i,j} = 0\),
    % 当 \(i \neq j\) 时,
    % 则 \(f(D) = [D]_{1,1} [D]_{2,2} \dots [D]_{n,n}\).

    我们说明, \(f(A) = \det {(A)}\),
    对任何 \(n\)~级阵 \(A\).

    设 \(A\) 是 \(n\)~级阵.
    设 \(p\) 是不超过 \(n\) 的正整数.
    设 \(s\) 是数.
    作 \(n\)~级阵 \(M(n; p; s)\) 如下:
    \begin{align*}
        [M(n; p; s)]_{i,j} =
        \begin{cases}
            [I_n]_{i,j},   & j \neq p; \\
            s [I_n]_{i,p}, & j = p.
        \end{cases}
    \end{align*}
    我们考虑 \(A M(n; p; s)\).
    当 \(j \neq p\) 时,
    \begin{align*}
        [A M(n; p; s)]_{i,j}
        = {} &
        \sum_{k = 1}^{n} {[A]_{i,k} [M(n; p; s)]_{k,j}}
        \\
        = {} &
        \sum_{k = 1}^{n} {[A]_{i,k} [I_n]_{k,j}}
        \\
        = {} &
        [A I_n]_{i,j}
        \\
        = {} &
        [A]_{i,j}.
    \end{align*}
    当 \(j = p\) 时,
    \begin{align*}
        [A M(n; p; s)]_{i,j}
        = {} &
        \sum_{k = 1}^{n} {[A]_{i,k} [M(n; p; s)]_{k,p}}
        \\
        = {} &
        \sum_{k = 1}^{n} {[A]_{i,k} s [I_n]_{k,p}}
        \\
        = {} &
        s [A I_n]_{i,p}
        \\
        = {} &
        s [A]_{i,p}.
    \end{align*}
    则 \(A M(n; p; s)\) 的列~\(j\) 等于 \(A\) 的列~\(j\)
    (\(j \neq p\)),
    且 \(A M(n; p; s)\) 的列~\(p\) 是 \(A\) 的列~\(p\)
    的 \(s\) 倍.
    注意, 若 \(i \neq j\), 则 \([M(n; p; s)]_{i,j} = 0\).
    则 \(f(M(n; p; s)) = \det {(M(n; p; s))} = s\).
    % 则 \(f(M(n; p; s)) = s\).
    则 \(f(A M(n; p; s)) = f(A) f(M(n; p; s))
    = s f(A)\).
    用 ``用行列式的性质确定行列式'' 的话,
    这说明, \(f\) 有多齐性.
    若我们还能说明 \(f\) 有倍加不变性,
    则, 由 \(f(I_n) = \det {(I_n)} = 1\)
    (注意, 若 \(i \neq j\), 则 \([I_n]_{i,j} = 0\)),
    % (这是显然的),
    我们立得 \(f(A) = \det {(A)}\).

    我们知道, 可用阵的积表示倍加.
    由 ``阵的积与倍加'' 的知识,
    对 \(n\)~级阵 \(A\),
    加 \(A\) 的列~\(p\) 的 \(s\)~倍于列~\(q\)
    (其中, \(p \neq q\)),
    且不改变别的列, 得 \(A E(n; p, q; s)\),
    其中, \(n\)~级阵 \(E(n; p, q; s)\) 的元
    \begin{align*}
        [E(n; p, q; s)]_{i,j}
        = \begin{cases}
              s,           & \text{\(i = p\), 且 \(j = q\)}; \\
              [I_n]_{i,j}, & \text{别的情形}.
          \end{cases}
    \end{align*}
    % (类似地,
    % 加 \(A\) 的行~\(q\) 的 \(s\)~倍于行~\(p\),
    % 且不改变别的行, 得 \(E(n; p, q; s) A\).)
    则 \(f(A E(n; p, q; s)) = f(A) f(E(n; p, q; s))\).
    我们的目标是, \(f(E(n; p, q; s)) = 1\).
    % Postscript on 2024-08-19:
    % If n ≥ 3, then
    % E(n; p, q; s) =
    % E(n; p, w; s) E(n; w, q; 1) E(n; p, w; -s) E(n; w, q; -1),
    % in which w ≠ p and w ≠ q and p ≠ q.
    % Commutators are useful, are they not.
    %
    % Alternatively, I have shown that
    % for t with t ≠ 0 and t ≠ 1,
    % M(n; p; t) =
    % E(n; p, q; -s/(1-t)) E(n; p, q; s) M(n; p; t) E(n; p, q; s/(1-t)),
    % or, equivalently,
    % E(n; p, q; s) =
    % E(n; p, q; s/(1-t)) M(n; p; t) E(n; p, q; -s/(1-t)) M(n; p; 1/t).

    设数 \(t \neq 1\), 且 \(t \neq 0\).
    记
    \(U = E(n; p, q; s/(1-t))\),
    \(V = E(n; p, q; -s/(1-t))\),
    \(G = M(n; p; t)\),
    且 \(H = M(n; p; 1/t)\).
    注意,
    \([UV]_{p,q} = [U]_{p,q} + [U]_{q,q}\, (-s/(1-t)) = 0\),
    且
    \([UV]_{i,j} = [U]_{i,j} = [I_n]_{i,j}\)
    (别的情形).
    故 \(UV = I_n\).
    注意,
    \([GH]_{p,p} = [G]_{p,p}\, (1/t) = 1\),
    且
    \([GH]_{i,j} = [G]_{i,j} = [I_n]_{i,j}\)
    (别的情形).
    故 \(GH = I_n\).
    我们计算 \(UGVH\)
    (注意, 这不是 \(UVGH\)).
    首先,
    \([UG]_{p,p} = [U]_{p,p}\, t = t\),
    \([UG]_{p,q} = [U]_{p,q} = s/(1-t)\),
    且
    \([UG]_{i,j} = [U]_{i,j}\)
    (别的情形).
    则
    \([UGV]_{p,p} = [UG]_{p,p} = t\),
    \([UGV]_{p,q} = [UG]_{p,q} + [UG]_{p,p}\, (-s/(1-t))
    = s/(1-t) - st/(1-t) = s\),
    且
    \([UGV]_{i,j} = [UG]_{i,j} = [U]_{i,j}\)
    (别的情形).
    则
    \([UGVH]_{p,p} = [UGV]_{p,p}\, (1/t) = 1\),
    \([UGVH]_{p,q} = [UGV]_{p,q} = s\),
    且
    \([UGVH]_{i,j} = [UGV]_{i,j} = [U]_{i,j}\)
    (别的情形).
    则 \(UGVH = E(n; p, q; s)\).
    故
    \begin{align*}
        f(E(n; p, q; s))
        = {} &
        f(UGVH)
        \\
        = {} &
        f(UG) f(VH)
        \\
        = {} &
        f(U) f(G) f(V) f(H)
        \\
        = {} &
        f(U) f(V) f(G) f(H)
        \\
        = {} &
        f(UV) f(GH)
        \\
        = {} &
        f(I_n) f(I_n)
        \\
        = {} &
        1.
    \end{align*}

    % https://www.quora.com/What-is-the-proof-that-the-matrix-determinant-is-uniquely-determined-by-the-rule-for-diagonal-matrices-and-det-AB-det-A-det-B-rules
    % 取不超过 \(n\) 的正整数 \(p\), \(q\),
    % 且 \(p \neq q\).
    % 取数 \(s\).
    % 设数 \(t \neq 1\), 且 \(t \neq 0\).
    % 记 \(L = E(n; p, q; s) M(n; p; t)\).
    % 则
    % \([L]_{p,q} = [M(n; p; t)]_{p,q}
    % + s [M(n; p; t)]_{q,q} = s\),
    % 且 \([L]_{i,j} = [M(n; p; t)]_{i,j}\),
    % 当 \(i \neq p\) 或 \(j \neq q\).
    % 我们用二个方式算 \(f(L)\).

    % 一方面,
    % \(f(L) = f(E(n; p, q; s)) f(M(n; p; t))
    % = f(E(n; p, q; s))\, t\).

    % 另一方面, 我们可倍加二次,
    % 变 \(L\) 为某个阵 \(D\),
    % 使 \(f(L) = f(D)\),
    % 且 \(i \neq j\) 时, \([D]_{i,j} = 0\).
    % 记 \(U = E(n; p, q; s/(1-t))\)
    % 与 \(V = E(n; p, q; -s/(1-t))\).
    % 则 \([LU]_{p,q}
    %     = [L]_{p,q} + [L]_{p,p} s/(1-t) = s/(1-t)\),
    % 且 \([LU]_{i,j} = [L]_{i,j}\),
    % 当 \(i \neq p\) 或 \(j \neq q\).
    % 进一步, \([VLU]_{p,q} = [V(LU)]_{p,q}
    %     = [LU]_{p,q} + [LU]_{q,q} (-s/(1-t)) = 0\),
    % 且 \([VLU]_{i,j} = [LU]_{i,j} = [L]_{i,j}\),
    % 当 \(i \neq p\) 或 \(j \neq q\).
    % 则 \(VLU = M(n; p; t)\).
    % 故 \(f(VLU) = t\).
    % 注意,
    % \begin{align*}
    %     f(VLU)
    %     = {} &
    %     f(V (LU)) = f(V) f(LU) = f(LU) f(V)
    %     \\
    %     = {} &
    %     f((LU) V) = f(L (UV)).
    % \end{align*}
    % 注意,
    % \([UV]_{p,q} = [U]_{p,q} + [U]_{q,q} (-s/(1-t)) = 0\),
    % 且
    % \([UV]_{i,j} = [U]_{i,j} = [I_n]_{i,j}\),
    % 当 \(i \neq p\) 或 \(j \neq q\).
    % 故 \(UV = I_n\).
    % 则 \(t = f(VLU) = f(L(UV)) = f(L I_n) = f(L)\).

    % 比较二次计算的结果, 我们有
    % \(t = f(L) = f(E(n; p, q; s))\, t\).
    % 因为 \(t \neq 0\),
    % 由消去律, \(f(E(n; p, q; s)) = 1\).
    % 则 \(f\) 有 ``倍加不变性''.

    % 综上, 我们有了新的用行列式的性质确定行列式的方式.
\end{example}
